The values into the equation for the span (L), the span
[tex]L = ((0.75 * 384 * E * I_H) / (5 * w_actual))^0.25[/tex]
To find the span of the H-beam that can support a load of 17,500 lbf with a deflection of 0.75 in and a safety factor of 1.75, we need to use the formula for beam deflection.
The formula for beam deflection is given by:
δ = (5 * w * L^4) / (384 * E * I)
where:
δ is the deflection
w is the load per unit length
L is the span of the beam
E is the modulus of elasticity
I is the moment of inertia
Since the beam is an H-beam, the moment of inertia (I) will be different from that of an I-beam. To calculate the moment of inertia for an H-beam, we need the dimensions of the beam's cross-section.
Assuming the dimensions of the H-beam cross-section are known, we can calculate the moment of inertia (I). Let's denote it as I_H.
Once we have the moment of inertia (I_H), we can rearrange the deflection formula to solve for the span (L):
L = ((δ * 384 * E * I_H) / (5 * w))^0.25
Given the load of 17,500 lbf and the deflection of 0.75 in, we can calculate the load per unit length (w) as:
w = 17,500 lbf / L
Using the safety factor of 1.75, we multiply the load per unit length by the safety factor to get the actual design load per unit length (w_actual):
w_actual = 1.75 * w
Finally, substituting the values into the equation for the span (L), we can solve for the span:
L = ((0.75 * 384 * E * I_H) / (5 * w_actual))^0.25
Please provide the dimensions of the H-beam cross-section (width, height, and thickness) and the modulus of elasticity (E) to calculate the span of the beam.
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a turtle is moving along a straight line at a constant rate. it moves 2/7 of a mile in 3/5 of an hour. what is the average speed of the turtle
Answer:
I am not sure , but i think the answer would be 0.03 m/s .
hope it helps u . ^.^
The degree of precision of a quadrature formula whose error term is 24 f'"'() is: 5 4 3 2
The degree of precision of the quadrature formula with an error term of 24 f‴() is 2.
To know more about the degree of precision of a quadrature formula, refer here:
The degree of precision of a quadrature formula represents the highest power of x that the formula can integrate exactly. In this case, the error term of the formula is given as 24 f‴(), where f‴() denotes the third derivative of the function f(x). The degree of precision is determined by the highest power of x that appears in the error term.
In a quadrature formula, the error term typically has the form K * h^p, where K is a constant, h is the step size, and p is the degree of precision. In this case, the error term is 24 f‴(). We can see that there is no dependence on the step size h, which implies that h^p = h^0 = 1. Therefore, the highest power of x in the error term is determined by the highest power of x that appears in f‴().
Since the error term is 24 f‴(), it indicates that the highest power of x in f‴() is 1. Thus, the degree of precision of the quadrature formula is 2, as the highest power of x in the error term is two degrees less than the highest power of x that the formula can integrate exactly.
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if h(x) = 5 4f(x) , where f(4) = 5 and f '(4) = 2, find h'(4).
The derivative h'(4) is equal to 40.
To find h'(4), we can apply the chain rule. The chain rule states that if we have a composite function, we can find the derivative by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
In this case, h(x) = 5 * 4f(x), so the outer function is multiplying by 5 and the inner function is f(x).
First, we find the derivative of the outer function, which is a constant multiple. The derivative of 5 is 0.
Next, we find the derivative of the inner function, f'(x). We are given that f'(4) = 2.
Now, we can apply the chain rule: h'(x) = 0 * f(x) + 5 * f'(x).
Substituting the given values, we have: h'(4) = 0 * f(4) + 5 * f'(4) = 0 * 5 + 5 * 2 = 0 + 10 = 10.
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Please help with this question?!?
Answer:
50.3
Step-by-step explanation:
area for circle formula=πr^2
r=4
π4^2 which is 50.3 rounded
Answer:
50.3
Step-by-step explanation:
A=πr^2
π·4^2≈50.3
help me understand please
Answer:
V=pi x (r)^2 x h
V-volume r-radius h-height
V= pi x (3)^2 x 9
V= pi x 9 x 9
V= pi x 81
V= 254.47 (rounded)
In cell B23, find the required sample size to estimate the true proportion of satisfied customers within 3% margin of error, with 99% confidence. in excel??
To calculate the required sample size in Excel for estimating the true proportion of satisfied customers within a 3% margin of error with 99% confidence,
In Excel, you can use the NORM.S.INV function to find the critical value corresponding to the desired confidence level. In this case, we use 1 - (1-0.99)/2 to find the z-score for 99% confidence level.
We square this z-score and multiply it by (0.50.5)/(0.030.03), which represents the maximum variance under worst-case scenario and the desired margin of error. The CEILING function is used to round up the result to the nearest whole number since the sample size should be an integer.
you can use the formula =CEILING(MROUND((NORM.S.INV(1-(1-0.99)/2,0)^2)(0.50.5)/(0.03*0.03),1),1). The result will give you the minimum sample size needed for the desired confidence level and margin of error.
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Need this answer as soon as possible
Answer:
x=22
Step-by-step explanation:
Rule states diagonals = same so 6x-36 = 96 in which x=22 cause 6(22)=132
132-36 = 96
find the value of two numbers if their sum is 39 and their difference is 1
Find the work done by the force field F in moving an object from P to Q.
F(x, y) = x^5i + y^5j; P(1, 0), Q(3, 3)
The work done by the force field F in moving an object from P to Q is 1459/6.
The work done by a force field in moving an object from point P to point Q can be calculated using the line integral of the force field along the path from P to Q.
Given the force field F(x, y) = x^5i + y^5j and the points P(1, 0) and Q(3, 3), we need to calculate the line integral of F along the path from P to Q.
The line integral of a vector field F along a curve C is given by the formula:
∫C F · dr,
where F is the vector field, dr is the differential vector along the curve, and ∫C represents the line integral over the curve C.
In this case, the path from P to Q can be parameterized by a function r(t) = (x(t), y(t)), where t varies from 0 to 1. We can choose a linear parameterization for simplicity:
x(t) = 1 + 2t,
y(t) = 3t,
where t varies from 0 to 1.
Now, we can calculate the line integral:
∫C F · dr = ∫₀¹ F(x(t), y(t)) · r'(t) dt,
where r'(t) represents the derivative of r(t) with respect to t.
Substituting the values into the formula, we have:
∫₀¹ ( (1 + 2t)^5i + (3t)^5j ) · (2i + 3j) dt.
Simplifying the expression, we get:
∫₀¹ ( (1 + 2t)^5(2) + (3t)^5(3) ) dt.
Now, we can integrate term by term:
= ∫₀¹ ( 2(1 + 2t)^5 + 3^5t^5 ) dt.
= [ (2/6)(1 + 2t)^6 + (3/6)t^6 ] from 0 to 1.
Evaluating the expression at the limits, we have:
= (2/6)(1 + 2(1))^6 + (3/6)(1)^6 - (2/6)(1 + 2(0))^6 - (3/6)(0)^6.
= (2/6)(3^6) + (3/6) - (2/6)(1^6) - (3/6)(0).
= (2/6)(729) + (3/6) - (2/6) - 0.
= (2/6)(729 - 1) + (3/6).
= (2/6)(728) + (3/6).
= 728/3 + 3/6.
= 728/3 + 1/2.
= (1456 + 3)/6.
= 1459/6.
Therefore, the work done by the force field F in moving an object from P to Q is 1459/6.
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A factory recently installed new pollution control equipment ("scrubbers") on its smokestacks in hopes of reducing air pollution levels at a nearby national park. Randomly timed measurements of sulfate levels (in micrograms per cubic meter) were taken before (Set C1) and after (Set C2) the installation. We believe that measurements of sulfate levels are normally distributed. Write a complete conclusion about the effectiveness of these scrubbers based on the statistical software printout shown. SET C1 10.0 8.0 8.0 7.0 6.0 9.0 11.5 8.0 9.5 7.5 5.0 10.0 5.0 7.0 1.0 9.0 1.5 5.0 25 4.0 9.0 6.0 SET C2 Two Sample T for C1 vs C2 N Mean StDev SEMean G 8.29 1.83 0.53 C2 10 5.00 2.84 0:90 95% CI for mul-mu2: (1.20, 5.38) T-Test mu1 = mu2 (vs. not =) P=0.0037 DF = 20 I T= 3.29
We can see from the statistical software printout that the factory recently installed new pollution control equipment ("scrubbers") on its smokestacks in hopes of reducing air pollution levels at a nearby national park.
Here, randomly timed measurements of sulfate levels (in micrograms per cubic meter) were taken before (Set C1) and after (Set C2) the installation. We believe that measurements of sulfate levels are normally distributed. A complete conclusion about the effectiveness of these scrubbers based on the statistical software printout shown is:The mean sulfate level for Set C1 is 8.29, and the mean sulfate level for Set C2 is 5.00.
From the Two Sample T for C1 vs C2 table, we find that the t-test statistic is 3.29, and the p-value is 0.0037. Because the p-value (0.0037) is less than the level of significance (α=0.05), we can reject the null hypothesis that the mean sulfate levels are equal before and after the installation of the scrubbers. Therefore, we can conclude that the installation of scrubbers on the factory's smokestacks has had a significant effect in reducing the sulfate levels in the air at the national park.
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Un paseo tiene una longitud de 1250 km. Se quiere plantar un árbol cada 0,025 km. ¿cuantos arboles se plantaran? ¿ y si se planta un arbol cada 50 m?
Answer:
-Se plantarán 50.000 arboles.
-Si se planta un árbol cada 50 m, se plantarán 25.000 arboles.
Step-by-step explanation:
Para encontrar cuántos arboles se plantarán, tienes que dividir la longitud el paseo entre la distancia que habrá entre los árboles:
1250/0,025=50.000
Ahora, para encontrar cuántos arboles se plantarán si se planta uno cada 50m, primero tienes que encontrar cuántos kilómetros hay en 50 m considerando que 1 km es igual a 1000 metros. Para esto, puedes utilizar una regla de tres:
1 km → 1000 m
x ← 50 m
x=(50*1)/1000
x= 0,05 km
Ahora puedes dividir la longitud del paseo entre 0,05 para encontrar el número de arboles que se plantarán:
1250/0.05=25.000
De acuerdo a esto, la respuesta es que se plantarán 50.000 arboles.
Si se planta un árbol cada 50 m, se plantarán 25.000 arboles.
Given the set { 1 , 1 , 1 , 1 , 2 , 3 } . Picking a number at random .
The probability of choosing a "1" is 2 / 3 .
True
False
Given the set { 1 , 1 , 1 , 1 , 2 , 3 } . Picking a number at random.The probability of choosing a "1" is 2 / 3 . False
In the given set {1, 1, 1, 1, 2, 3}, there are four occurrences of the number "1" and a total of six elements in the set.
To calculate the probability of choosing a "1", we need to divide the number of favorable outcomes (the occurrences of "1") by the total number of possible outcomes (the total number of elements in the set).
The number of favorable outcomes is 4 (the occurrences of "1"), and the total number of possible outcomes is 6 (the total number of elements in the set).
So, the probability of choosing a "1" is 4/6, which simplifies to 2/3.
Therefore, the statement is true: the probability of choosing a "1" from the given set is 2/3.
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Olivia is preparing many gifts for a family holiday party. Each gift must either be wrapped with wrapping paper or prepared in a gift bag. Olivia can wrap a single gift in 3 minutes and she can do a giftbag in 1.5 minutes. Olivia prepared a total 14 gifts in 30 minutes. Graphically solve a system of equations in order to determine the number of gifts Olivia wrapped, x,x, and the number of giftbags she prepared, yy.
Answer:
Olivia wrapped 6 gifts and prepared 8 giftbags.
Step-by-step explanation:
Write System of Equations:
3x+1.5y= 30 x+y=14
x+y= 14 y=-x+14
3x+1.5y=30
1.5y=3x+30
1.5y divided by 1.5 is y
-3x+30 divided by y is -2x+20
y=-2x+20
Olivia wrapped 6 gifts and made 8 Giftbags.
What is Equation?Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.
It demonstrates the equality of the relationship between the expressions printed on the left and right sides. LHS = RHS is a common mathematical formula.
Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.
Given:
Olivia can wrap a single gift in 3 minutes and she can do a giftbag in 1.5 minutes
Olivia prepared a total 14 gifts in 30 minutes.
let number of gifts Olivia wrapped be x.
and, number of giftbags she prepared be y.
So, the equations are:
3x+1.5y= 30 .....(1)
x+y=14
Now, plotting these equations on graph we get intersecting point of two lines which shows x= 6 and y=8.
Hence, Olivia wrapped 6 gifts and made 8 Giftbags.
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On a coordinate plane, a curved line with a minimum value of (1.5, negative 1) and a maximum value of (negative 1.5, 13), crosses the x-axis at (negative 3, 0), (1, 0), and (2, 0), and crosses the y-axis at (0, 6).
Which lists all of the x-intercepts of the graphed function?
(0,6)
(1,0)(2,0)
(1,0)(2,0) and (-3,0)
(1,0)(2,0)(-3,0) and (0,6)
Answer:
c
Step-by-step explanation:
Can someone help me with this. Will Mark brainliest.
Answer:
1.8
Step-by-step explanation:
Solve this please!!!!
Answer:
The volume of the rectangular prism is 336
The volume of the triangular prism is 24.875
Add them together, 360.875. You can round if you want
In the diagram, mBDA = 150°. Find mBDC.
Answer:
a. 70°
Step-by-step explanation:
m∠BDC + m∠CDA = m∠BDA
-3x + 34 - 2x + 56 = 150
-5x + 90 = 150
-5x = 60
x = -12
m∠BDC = -3x + 34 = -3(-12) + 34
= 36 + 34 = 70°
Plz help will mark brainliest if right
Answer:
Domain:{-9,-4}
Range:{-1,-8,6}
Step-by-step explanation:
I dont really understand what you're trying to ask, it seems like you already have the answer i got.
Please help:///. Due today
Answer:
C: 108 cans
Step-by-step explanation:
just trust me bro
Margarita was buying soil for her spring garden. What would be the most reasonable amount of soil she would need to buy for a garden?
A. 5 grams
B. 50 liters
C. 50 kilograms
D. 5,000 kilograms
Answer:
C. 50 kilograms
Step-by-step explanation:
For a normal house garden of moderate dimensions 5,000 kg is too much and 50 kg would be enough.
Answer: I'm hoping for it to be C.
Step-by-step explanation: I'll finish the quiz then l'll let you know ;)
Edit: GOT IT CORRECT!
And don't mind that i got three wrong...
In an experiment a bag contains 2 blue marbles and 5 red marbles. Two marbles are drawn from the bag.
Answer:
5
Step-by-step explanation:
2-5=5. Subscribe Single-3 FF on yt
The sample space of all possible pairs of marbles that can be drawn is:
{BB, BR, RB, RR, RR, RR}
What is sample space?It is the total number of possible outcomes from a given set.
We have,
The sample space is the set of all possible outcomes of the experiment.
In this case, we are drawing two marbles from a bag containing 2 blue and 5 red marbles, without replacement (meaning that we do not put the first marble back into the bag before drawing the second one).
The sample space consists of all possible pairs of marbles that can be drawn:
{BB, BR, RB, RR, RR, RR}
where BB means both marbles are blue, BR means the first marble is blue and the second is red, RB means the first marble is red and the second is blue, and RR means both marbles are red.
Note that there are two outcomes corresponding to drawing two red marbles because there are 5 red marbles in the bag.
Thus,
The sample space of all possible pairs of marbles that can be drawn is:
{BB, BR, RB, RR, RR, RR}
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The complete question:
In an experiment, a bag contains 2 blue marbles and 5 red marbles. Two marbles are drawn from the bag.
List the sample space.
A librarian noticed that 14% of the seventh graders checked out science fiction books and 12% checked out sports book. Out of 350 seventh graders, how many students should she expect NOT to check out science fiction and sports books?
Answer:
259 students
Step-by-step explanation:
First, find how many will check out science fiction books
350(0.14)
= 49
Find how many will check out sports books
350(0.12)
= 42
Subtract these amounts from 350
350 - 49 - 42
= 259
So, she should expect 259 students not to check out science fiction and sports books
Suppose that {Y, }is stationary with mean 3 and autocovariance function y(k). Let Couche, "t-m) R = 2Y - Y-1 Find the mean and autocovariance function of Rt. Is R stationary? Why? ry THE 1 2 7u
The mean of Rt can be calculated as follows:
E[Rt] = E[2Yt - Yt-1] = 2E[Yt] - E[Yt-1] = 2(3) - 3 = 3.
and, the autocovariance function of Rt is given by:
yR(k) = 4y(k) - 2y(k+1) - 2y(k-1) + y(k).
The random variable Rt is defined as 2Yt - Yt-1, where {Yt} is a stationary process with a mean of 3 and autocovariance function y(k). To find the mean and autocovariance function of Rt, we can use the properties of linearity and time shift.
The mean of Rt can be calculated as follows:
E[Rt] = E[2Yt - Yt-1] = 2E[Yt] - E[Yt-1] = 2(3) - 3 = 3.
To find the autocovariance function of Rt, we need to consider the covariance between Rt at time t and Rt at time s for any t and s. Let's denote the autocovariance function of Yt as γ(k). Using the properties of linearity and time shift, we can express the autocovariance function of Rt as follows:
Cov(Rt, Rs) = Cov(2Yt - Yt-1, 2Ys - Ys-1)
= 4Cov(Yt, Ys) - 2Cov(Yt, Ys-1) - 2Cov(Yt-1, Ys) + Cov(Yt-1, Ys-1)
= 4γ(t-s) - 2γ(t-s+1) - 2γ(t-1-s) + γ(t-1-s+1)
= 4γ(t-s) - 2γ(t-s+1) - 2γ(t-s-1) + γ(t-s).
Therefore, the autocovariance function of Rt is given by:
yR(k) = 4y(k) - 2y(k+1) - 2y(k-1) + y(k).
Now, let's determine whether R is stationary. A process is considered stationary if its mean and autocovariance do not depend on time. In this case, the mean of Rt is constant and equal to 3, which satisfies the condition for stationarity. However, the autocovariance function yR(k) of Rt depends on the lag k. Therefore, Rt is not stationary because its autocovariance function varies with time.
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For n e N let an be the number of strings of 0's and l's such that every 0 is followed by a 1. (a) Write down the values for a¡ through as (you may need to use some scratch paper) i. ai = ii. 02 = iii. a3 = iv. 04 = V. a5 = (b) Do these numbers look familiar? Make a conjecture. (C) How might you build the set of these strings of length n from the sets of strings of length n - 1 and strings of length n - 2? (d) Prove your conjecture.
For n e N let an be the number of strings of 0's and l's,
(a) i. a₁ = 2, ii. a₂ = 1, iii. a₃ = 2, iv. a₄ = 3, v. a₅ = 5.
(b) The values of aₙ represent the nth term in the Fibonacci sequence.
(c) The set of strings of length n can be built by appending "1" or "01" to strings of length n-1 and n-2.
(d) The conjecture that the values of aₙ represent the nth term in the Fibonacci sequence is proven using mathematical induction.
(a) To find the values of aₙ, we'll calculate them one by one:
i. a₁: We have two possible strings of length 1 that satisfy the condition: "1" and "0". So a₁ = 2.
ii. a₂: For a string of length 2, the only valid option is "10". So a₂ = 1.
iii. a₃: Now, let's consider strings of length 3. We can build them by appending either "1" or "01" to valid strings of length 2. From the previous step, we know that a₂ = 1. Thus, we have two options: "101" and "100". So a₃ = 2.
iv. a₄: Similarly, for strings of length 4, we can append either "1" or "01" to valid strings of length 3. From the previous step, we know that a₃ = 2. Thus, we have three options: "1010", "1001", and "1000". So a₄ = 3.
v. a₅: Continuing the same pattern, we can append either "1" or "01" to valid strings of length 4. From the previous step, we know that a₄ = 3. Thus, we have five options: "10101", "10100", "10010", "10001", and "10000". So a₅ = 5.
(b) If we look closely at the values of aₙ that we calculated, we notice that they form the Fibonacci sequence: 2, 1, 2, 3, 5, ...
Conjecture: The values of aₙ represent the nth term in the Fibonacci sequence.
(c) To build the set of strings of length n from the sets of strings of length n-1 and n-2, we can append either "1" or "01" to the strings of length n-1 and n-2.
For example, to obtain a string of length 4, we can append "1" or "01" to the strings of length 3. Similarly, to obtain a string of length 5, we can append "1" or "01" to the strings of length 4.
(d) To prove our conjecture that the values of aₙ represent the nth term in the Fibonacci sequence, we can use mathematical induction. We would need to show that a₁ = F₁, a₂ = F₂, and assume that aₖ = Fₖ and aₖ₋₁ = Fₖ₋₁ hold for some k ≥ 2 and prove that aₖ₊₁ = Fₖ₊₁.
Since a₁ = 2 = F₁ and a₂ = 1 = F₂, the base cases hold.
Next, assume that aₖ = Fₖ and aₖ₋₁ = Fₖ₋₁ hold for some k ≥ 2.
From the construction in part (c), we know that to obtain a string of length k+1, we append "1" to a string of length k and append "01" to a string of length k-1.
So, aₖ₊₁ = aₖ + aₖ₋₁ = Fₖ + Fₖ₋₁.
Using the property of the Fibonacci sequence that Fₖ + Fₖ₋₁ = Fₖ₊₁, we can conclude that aₖ₊₁ = Fₖ₊₁.
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Last summer Olivia earned $3800 by selling cakes. She invested some of the money in a savings account that paid 4.5%/yr and the rest in a government bond that paid 5.5%/yr. After one year, she has earned $180 in interest. How much did she invest at each rate? [4]
Answer:
Olivia invested $2900 in a savings account that paid 4.5%/yr and $900 in a government bond that paid 5.5%/yr.
Step-by-step explanation:
With the information provided you can write the following equations:
x+y=3800 (1)
0.045x+0.055y=180 (2), where:
x is the amount invested in a savings account
y is the amount invested in a government bond
First, you can solve for x in (1):
x=3800-y (3)
Next, you can replace (3) in (2) and solve for y:
0.045(3800-y)+0.055y=180
171-0.045y+0.055y=180
0.01y=9
y=9/0.01
y=900
Finally, you can replace the value of y in (3) to find x:
x=3800-900
x=2900
According to this, the answer is that Olivia invested $2900 in a savings account that paid 4.5%/yr and $900 in a government bond that paid 5.5%/yr.
A scientist is studying the effect of rainfall on the height of a certain species of sunflower. The scatter plots shows the results.
Answer:answer is d
Step-by-step explanation:
I have done the test
Solve the system by substitution: 3x - y = 21 and y = 6x
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Answer:
if y=6x then 3x-6x=21
-3x=21
x=-7
3x - y = 21
y = 6x
We see that,
0 = 42 which is not true.
The system of equations has no solutions.
What is an equation?An equation is a mathematical statement that is made up of two expressions connected by an equal sign.
We have,
System of equations:
Conditions:
1 - If we get x = n (a value)
The system of equations has one solution
2 - If we get x = x or y = y or n = n
The system of equations has infinite solutions.
3 - If we get n ≠ 0
The system of equations has no solutions
Now,
3x - y = 21 _____(1)
y = 6x ______(2)
From (1) we get,
3x = 21 + y
x = (21 + y) / 3 _____(3)
Putting (3) in (2) we get,
y = 6 [(21 + y) / 3]
y = 2 (21 + y)
y = 42 + y
0 = 42
Which is not true.
This means the system of equations has no solutions.
Thus,
The system of equations has no solutions.
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If X has an exponential distribution, which of the following statements is correct?
A. The exponential distribution has the property that its mean equals its variance. B. All of the given statements are correct. C. The exponential distribution is governed by two parameters that determine its shape and location. D. The cumulative density function for an exponential random variable x has a bell-shaped graph. E. The exponential distribution is sometimes called the waiting-time distribution, because it is used to describe the length of time between occurrences of random events.
option A is the correct statement regarding the exponential distribution.
What is Exponential Distribution?
The exponential distribution is a continuous probability distribution that models the time between events occurring at a constant average rate. It is often used to describe situations where events happen randomly and independently over time, such as the time between phone calls received at a call center or the time between arrivals of customers at a service counter.
The correct answer is:
[tex]\textbf{A. The exponential distribution has the property that its mean equals its variance.}[/tex]
The exponential distribution is characterized by the property that its mean is equal to its variance. This means that the average value of the distribution and the spread of the distribution are the same. In other words, the measure of central tendency and the measure of dispersion are equal in the exponential distribution.
This property holds true for all exponential distributions, regardless of the specific rate parameter or scale parameter. It is a fundamental characteristic of this distribution.
Therefore, option A is the correct statement regarding the exponential distribution.
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. Amy Company sold merchandise of $8,000 to Tory Turnbull with terms 2/10, n/30. Amy Company recorded this transaction using the gross method. If Tory Turnbull paid for all the merchandize within the discount period, the journal entry that Amy Company will make to record the collection of cash would include a: a. Credit to Sales Discount of $160 b. Credit to Account receivable $7,840 c. Debit to Sales Discount of $160 d. Credit to Cash of $160 Select-
The journal entry that Amy Company will make to record the collection of cash from Tory Turnbull, who paid within the discount period, would include a credit to Cash for $160. Therefore, option d is the correct answer.
This is because Tory Turnbull will pay $8,000 - $160 (2% of $8,000) to avail the discount. The Sales Discount account is not involved in the journal entry as the discount was taken by the customer, not given by Amy Company.
Therefore, the correct answer is d. Credit to Cash of $160. This entry reflects the cash received by Amy Company and the reduction in the Accounts Receivable balance for the amount paid by the customer.
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what is true about a rhombus and square?
A. all sides are congruent to each other
B. all of these are true
C. both have diagonals that are perpendicular to each other
D. both have diagonals that bisect each other
Answer:
B. all of these are true
Step-by-step explanation:
The Square is a rhombus because all the sides of a square are equal in length. Even, the diagonals of both square and rhombus are perpendicular to each other and bisect the opposite angles. Therefore, we can say the square is a rhombus.
Hope this helped!!