The American Hospital Association stated in its annual report that the mean cost to community hospitals per patient per day in U.S. hospitals was $1231 in 2007. In that same year, a random sample of 25 daily costs in the state of Utah hospitals yielded a mean of $1103. Assuming a population standard deviation of $252 for all Utah hospitals, do the data provide sufficient evidence to conclude that in 2007 the mean cost in Utah hospitals is below the national mean of $1231? Perform the required hypothesis test at the 5% significance level.

Answers

Answer 1

We can conclude that the null hypothesis is rejected. There is sufficient evidence to support the claim that the mean cost in Utah hospitals is below the national mean of $1231.

How is this so?

H₀: μ ≥ 1231 (The mean cost in Utah hospitals is greater than or equal to the national mean)

Hₐ: μ < 1231 (The mean cost in Utah hospitals is below the national mean)

Given

Sample mean (x) = $1103Sample size (n) = 25Population standard deviation (σ) = $252Significance level (α) = 0.05

The test statistic for a one-sample t-test is given by

t = (x - μ) / (σ / √n)

Substituting we have

t = (1103 - 1231) / (252 / √25)

≈ -6.103

To determine the critical value, we need to find the critical t-value at the 5% significance level with degrees of freedom

(df) equal to (n - 1)

= (25 - 1)

= 24.

Using a t-distribution table or calculator, the critical value is approximately -1.711.

Since the calculated test statistic (-6.103) is smaller than the critical value (-1.711) and falls into the critical region, we reject the null hypothesis.

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Related Questions

What is the lyrics for its good to be alive by among us?

Answers

Answer:

HEHEHEHEHEH

I've been waiting for this moment

Feels good to be alive right about now

Good, good, good, good to be alive right about now

Good, good, good, good to be alive right about now

Hallelujah, let that bass line move ya, say hey

Step-by-step explanation:

Have fun

A solid object has the right triangle with vertices (0, 0), (3, 0), and (0, 4) as its base.

a) Any cross section of the solid, taken parallel to the y-axis and perpendicular to the x-axis, is a square. Find the volume of the solid.
b) Any cross section of the solid, taken parallel to the y-axis and perpendicular to the x -axis, is a smi-circle. Find the volume of the solid.

Answers

a. The volume of the solid is 24 cubic units.

b. The volume of the solid is 4π cubic units.

How to calculate tie value

a. Volume = Area of Base * Height

The base is a right triangle with base length of 3 units and height of 4 units. The area of the base can be calculated as:

Area of Base = (1/2) * base * height

= (1/2) * 3 * 4

= 6 square units

The height of the solid is 4 units.

Volume = Area of Base * Height

= 6 * 4

= 24 cubic units

b) Any cross section of the solid, taken parallel to the y-axis and perpendicular to the x-axis, is a semicircle.

Volume = (1/2) * π * radius² × height

Volume = (1/2) * (1/2) * π * 2² * 4

= (1/4) * π * 4 * 4

= π * 4

Therefore, the volume of the solid is 4π cubic units.

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Solve the following initial value problem. cos2x sin x dy + (cos?x)y = 5, 7(7/3) = 4 dx

Answers

The solution to the given initial value problem [tex]cos2x sin x dy + (cos?x)y = 5, 7(7/3) = 4 dx[/tex]  is [tex]y(x) = (5/6) * (cos^2(x) - 1) + (28/9) * sin(x) + (40/9) * cos(x)[/tex].

To solve the initial value problem, we can use the method of integrating factors. The integrating factor is found by taking the exponential of the integral of the coefficient of y with respect to x. In this case, the coefficient is cos(x). So, the integrating factor is [tex]e^(^\int ^{ cos(x)} \, ^d^x^) = e^s^i^n^(^x^))[/tex].

Multiplying the given differential equation by the integrating factor, we obtain:

[tex]e^{sin(x)} * [cos^{2x} sin(x) dy + cos(x) y] = 5e^{sin(x)} dx[/tex]

By using the product rule and simplifying, we have:

[tex]d/dx [e^{sin(x)} * y * cos^{2x}] = 5e^{sin(x)}[/tex]

Integrating both sides with respect to x, we get:

[tex]e^{sin(x)} * y * cos^{2(x)} = \int {5e^{sin(x)}} \, dx = 5e^{sin(x)} + C,[/tex]

where C is the constant of integration.

Simplifying and solving for y, we obtain:

[tex]y(x) = [(5/6) * (cos^{2x} - 1) + (28/9) * sin(x) + (40/9) * cos(x)] / (e^{sin(x)} * cos^{2(x)} + Ce^{-sin(x)},[/tex]

where C is the constant of integration.

To find the value of C, we can use the initial condition y(7/3) = 4. Substituting this into the equation and solving for C, we can determine the specific solution.

In conclusion, the solution to the given initial value problem is [tex]y(x) = (5/6) * (cos^{2x} - 1) + (28/9) * sin(x) + (40/9) * cos(x)[/tex], subject to the specified initial condition.

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Write a quadratic equation in standard form with [tex]\frac{3}{4}[/tex] an -5 as its roots

Answers

Knowing the roots first write the equation in factored form:

(X - 3/4)(x +5) = 0

Now use the FOIL method ( multiply each term in one set of parentheses by each term the other set:

X•x + x•5 -3/4•x -3/4•5

Simplify:

X^2 + 5x -3/4x -3 3/4

Combine like terms:

X^2+ 4 1/4x - 3 3/4

select the correct answer. which expression is equivalent to x 3x2−2x−3÷x2 2x−3x 1 if no denominator equals zero? a. 1x2−2x−3 b. 1x2−4x 3 c. 1x2 2x−3 d. x 3x 1

Answers

The correct answer is option c. 1/(x² + 2x - 3). To determine which expression is equivalent to the given expression, let's simplify it step by step:

The given expression is (x³ - 2x - 3) ÷ (x² + 2x - 3).

Option a. 1/(x² - 2x - 3):

This option is not equivalent to the given expression because it represents the reciprocal of the quadratic denominator, which is different from the given expression.

Option b. 1/(x² - 4x + 3):

This option is not equivalent to the given expression because the signs of the quadratic terms are different. The given expression has a positive quadratic term, while this option has a negative quadratic term.

Option c. 1/(x² + 2x - 3):

This option is equivalent to the given expression because it represents the reciprocal of the quadratic denominator with the same signs for the quadratic terms.

Option d. x/(3x - 1):

This option is not equivalent to the given expression because it lacks the term x³ in the numerator.

Therefore, the correct answer is option c. 1/(x² + 2x - 3).

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Find the flux of the given vector field F across the upper hemisphere x^2 + y^2 + z^2 = a^2, z >= 0. Orient the hemisphere with an upward-pointing normal.
19. F= yj
20. F = yi - xj
21. F= -yi+xj-k
22. F = x^2i + xyj+xzk

Answers

6πa² is the flux of F across the upper hemisphere.

The problem requires us to compute the flux of the given vector field F across the upper hemisphere x² + y² + z² = a², z ≥ 0. We are to orient the hemisphere with an upward-pointing normal. The four vector fields are:

F = yj

F = yi - xj

F = -yi + xj - kz

F = x²i + xyj + xzk

To begin with, we'll make use of the Divergence Theorem, which states that the flux of a vector field F across a closed surface S is equivalent to the volume integral of the divergence of the vector field over the region enclosed by the surface, V, that is:

F · n dS = ∭V (div F) dV

where n is the outward pointing normal unit vector at each point of the surface S, and div F is the divergence of F.

We'll need to write the vector fields in terms of i, j, and k before we can compute their divergence. Let's start with the first vector field:

F = yj

We can rewrite this as:

F = 0i + yj + 0k

Then, we compute the divergence of F:

div F = d/dx (0) + d/dy (y) + d/dz (0)

= 0 + 0 + 0 = 0

So, the flux of F across the upper hemisphere is 0. Now, let's move onto the second vector field:

F = yi - xj

We can rewrite this as:

F = xi + (-xj) + 0k

Then, we compute the divergence of F:

div F = d/dx (x) + d/dy (-x) + d/dz (0)

= 1 - 1 + 0 = 0

So, the flux of F across the upper hemisphere is 0. Let's move onto the third vector field:

F = -yi + xj - kz

We can rewrite this as:

F = xi + y(-1j) + (-1)k

Then, we compute the divergence of F:

div F = d/dx (x) + d/dy (y(-1)) + d/dz (-1)

= 1 - 1 + 0 = 0

So, the flux of F across the upper hemisphere is 0. Lastly, let's consider the fourth vector field:

F = x²i + xyj + xzk

We can compute the divergence of F directly:

div F = d/dx (x²) + d/dy (xy) + d/dz (xz)

= 2x + x + 0 = 3x

Then, we express the surface as a function of spherical coordinates:

r = a, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2

Note that the upper hemisphere corresponds to 0 ≤ φ ≤ π/2.

We can compute the flux of F over the hemisphere by computing the volume integral of the divergence of F over the region V that is enclosed by the surface:

r² sin φ dr dφ dθ

= ∫[0,2π] ∫[0,π/2] ∫[0,a] 3r cos φ dr dφ dθ

= ∫[0,2π] ∫[0,π/2] (3a²/2) sin φ dφ dθ

= (3a²/2) ∫[0,2π] ∫[0,π/2] sin φ dφ dθ

= (3a²/2) [2π] [2] = 6πa²

Therefore, the flux of F across the upper hemisphere is 6πa².

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help with questions 10-13 plz!!

Answers

Answer:

._. ;

._. ;

._. ;

Could someone please help me with this !

And also show work

Answers

Answer: C. K= 2.5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Answer:

[tex]\Huge\boxed{k=2.5}[/tex]

Step-by-step explanation:

Given -2.1k + 13 + 6.5k = 24 we need to isolate the variable using inverse operations

step 1 combine any like terms

sometimes there are not like terms but in this case there are. When there are like terms (must be on the same side of the = ) you add them together

-2.1k + 6.5k = 4.4k

now we have

4.4k + 13 = 24

Now we want to get rid of the 13

To do so we subtract 13 from each side

13 - 13 cancels out

24 - 13 = 11

now we have 4.4k=11

now we want to get rid of the 4.4

To do so we divide each side by 4.4

4.4k/4=k

11/4.4=2.5

we're left with k - 2.5

An engine additive is being tested to see whether it can effectively increase gas mileage for a number of vehicles. Twenty assorted vehicles had their gas mileage, in miles per gallon, measured. Then, the engine additive was placed into each of the engines, and the gas mileage was measured again. Let

Answers

The calculated t-value is greater than 1.734, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

An engine additive is being tested to see whether it can effectively increase gas mileage for a number of vehicles. Twenty assorted vehicles had their gas mileage, in miles per gallon, measured. Then, the engine additive was placed into each of the engines, and the gas mileage was measured again. Let µd be the true mean difference in the gas mileage before and after the engine additive is placed into the engines. We want to test the hypothesis that the engine additive has no effect. The null hypothesis is: H0: µd = 0The alternative hypothesis is: Ha: µd > 0 (one-tailed test)Assuming that the difference in gas mileage before and after the engine additive is approximately normally distributed, we can use a one-sample t-test. The test statistic is given by: $$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$$Where, $\bar{x}$ is the sample mean difference in gas mileage, μ is the hypothesized population mean difference, s is the sample standard deviation of the differences, and n is the sample size. We can use a significance level of α = 0.05.To determine the critical value of the t-distribution, we need to find the degrees of freedom (df). Since we have a sample size of n = 20, we have n - 1 = 19 degrees of freedom. Using a t-distribution table with 19 degrees of freedom and a significance level of 0.05 for a one-tailed test, we get a critical value of t = 1.734. If the calculated t-value is greater than 1.734, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

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NOLINKS ..................

Answers

Answer:

The answer is in the link

Step-by-step explanation:

quntyfcjb/crown!.com :))))

Ok so use photo math instead

who are these bots giving us links i’m literally gonna fail now

Answers

Answer:

Fr. It’s annoying. My mom yells at me for not passing lolz.

Step-by-step explanation:

Answer:

Ignore them and re-upload the questions. The links are very bad.

Step-by-step explanation:

PLEASE HELP ASAP
Find the surface area and volume of the figure below:

Answers

Answer:

What I think it might be is 62.3

can i get an owa owa ?????

Answers

owa owa???

owa owa??

OWA OWA??

OWA OWA??

OWA OWA?

Find all possible trigonometric ratios given the following:

tan θ = -7/24 and cos θ > 0

Answers

The given information allows us to find the values of trigonometric ratios involving angle θ. Given that tan θ = -7/24 and cos θ > 0, we can determine the following trigonometric ratios: sin θ, csc θ, sec θ, and cot θ

We are given that tan θ = -7/24. Using this information, we can determine the values of sin θ and csc θ.

Since tan θ = sin θ / cos θ, we can write -7/24 = sin θ / cos θ. Rearranging the equation, sin θ = -7 and cos θ = 24.

Now, we can find the values of csc θ, sec θ, and cot θ.

csc θ is the reciprocal of sin θ, so csc θ = 1 / sin θ = 1 / (-7) = -1/7.

To find sec θ, we use the fact that sec θ = 1 / cos θ. So, sec θ = 1 / (24) = 1/24.

Lastly, to calculate cot θ, we know that cot θ = 1 / tan θ. Thus, cot θ = 1 / (-7/24) = -24/7.

In summary, given tan θ = -7/24 and cos θ > 0, we have sin θ = -7, csc θ = -1/7, sec θ = 1/24, and cot θ = -24/7.

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Find the distance between the points (7,

9) and (

2,

4).

Answers

Answer:

7.07106781187

Step-by-step explanation:

Let us use the  distance formula:

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}[/tex]

x2=2

x1=7

y2=4

y1=9

[tex]d = \sqrt{(2 - 7)^{2} + (4-9)^2}[/tex]

[tex]d = \sqrt{(-5)^2 + (-5)^2}[/tex]

[tex]d = \sqrt{25 + 25}[/tex]

[tex]d = \sqrt{50}[/tex]

d=7.07106781187 (round to whatever digit neccesary)

Hope this helps!!

how do you get the value of x7+8x+3-1+7

Answers

Answer:

first add the numbers that have the same variable s x7+8x=15x

then the numbers first add and then you subtract

1+7=8

3-8= -5

15x-5

Answer:

8x+x7+9

Step-by-step explanation:

subtracted one term from another

subtracted 1 from 3 to get 2

x7=+8x+2+7

add two terms together

add 2 and 7 toget 9

x7+8x+9

1. Identify the Parent function related to the given function. Choose the correct answer from the choices below:
f(x)=1/2x-9 4/5
Linear Function
Not a Function
Absolute Value Function
Quadratic Function

Answers

I believe it’s linear because it’s in y=mx+b form

I need help on the circled problem please

Answers

Answer:

There would be infinite solutions because the equations are exactly the same.

I hope this answered your question

THIS WILL HELP A LOT OF PPL PLZ HLP!!!!!!
Determine the interval where the graph of the function is negative.

ANSWER CHOICES AND GRAPH IN IMAGES

Answers

Answer:

If I'm correct I think its answer B

Step-by-step explanation:

I'm not sure but i hope this help

Answer:

second option

-∞ < x < 1

Step-by-step explanation:

How many permutations of S9, have cycle strucrure 3^3?

Answers

There is only 1 permutation in S9 with a cycle structure of [tex]3^3[/tex].

To find the number of permutations of S9 with a cycle structure of [tex]3^3[/tex], we can use the concept of cycle index.

In a permutation with a cycle structure of[tex]3^3[/tex], we have three cycles of length 3. The cycle index of S9 with respect to cycles of length 3 can be determined using the Polya enumeration theorem.

The cycle index of S9 with respect to cycles of length 3 is given by:

[tex]Z(S9, t1, t2, t3) = (t1^3 + t3^3)^3[/tex]

Expanding this expression, we get:

[tex]Z(S9, t1, t2, t3) = (t1^3 + t3^3)^3\\\= (t1^9 + 3t1^6t3^3 + 3t1^3t3^6 + t3^9)[/tex]

To count the number of permutations with the desired cycle structure, we need to find the coefficient of the term [tex]t1^9t3^9[/tex].

From the expanded form, we see that the coefficient  [tex]t1^9t3^9[/tex] is 1.

Therefore, there is only one permutation in S9 with a cycle structure of [tex]3^3[/tex]

In summary, there is 1 permutation of S9 that has a cycle structure of [tex]3^3[/tex].

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Show that if U is open in X, and A is closed in X, then UA is open in X, and A\U is closed in X.

Answers

The intersection of N(x) and N'(x), denoted by N(x)∩N'(x), is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.

Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.

To show that if U is open in X and A is closed in X, then UA is open in X and A\U is closed in X, we need to prove two statements:

UA is open in X.A\U is closed in X.

Let's prove these statements one by one:

To show that UA is open in X, we need to prove that for every point x in UA, there exists an open neighborhood around x that is completely contained within UA.

Let x be an arbitrary point in UA. Since x is in UA, it must belong to U as well as A. Since U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, since x is in A, it is also in X\U (complement of U in X). As A is closed in X, X\U is closed in X, which means its complement, U, is open in X. Therefore, there exists an open neighborhood N'(x) of x that is completely contained within X\U.

Now, consider the intersection of N(x) and N'(x), denoted by N(x)∩N'(x). This intersection is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.

Since N(x)∩N'(x) is an open neighborhood of x completely contained within UA, we have shown that UA is open in X.

To show that A\U is closed in X, we need to prove that its complement, (A\U)', is open in X.

Let x be an arbitrary point in (A\U)'. Since x is not in A\U, it means that x must either be in A or in U (or both). If x is in A, then x is not in A\U. Therefore, x is in U.

Since x is in U and U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, consider the intersection of N(x) and A. Since x is in A, N(x)∩A is a non-empty set. Let y be any point in N(x)∩A.

We know that N(x)∩A ⊆ U∩A ⊆ A\U, because if y was in U, it would contradict the assumption that y is in A. Therefore, N(x)∩A is a subset of A\U.

Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.

Hence, we have shown that if U is open in X and A is closed in X, then UA is open in X, and A\U is closed in X.

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Which golf ball went higher, and how many feet? (Desmos!)

Answers

Answer:

1. 36

2. Second

Step-by-step explanation:

- For the first ball, we can see the given function:

[tex]f(x)=-16(t^{2}-3t )[/tex]

[tex]=-16[t^{2} -3t+(3/2)^{2}-(3/2)^{2} ][/tex]

[tex]=-16(t-\frac{3}{2} )^{2} +(-\frac{3}{2} )^{2} *(-16)[/tex]

[tex]-16(t-\frac{3}{2} )^{2} +36[/tex]

So the vertex is ([tex]\frac{3}{2}[/tex], 36), it means when the ball was hit by the [tex]\frac{3}{2}[/tex] seconds, it arrived at the highest height of 36 feet.

- For the second ball, we can see the given graph: the vertex is (2,64), it means when the ball was hit by the 2 seconds, it arrived at the highest height of 64 feet.

- Compare to the two heights, 36 (first ball) is less than 64 (second ball), so the second ball went higher.

please someone answer fast!!! I'm so confused and this is due today

Answers

Answer:

From greatest to least it would be 3.66666,[tex]\sqrt{11}[/tex],2(1/4),-2.5,-3.97621

Step-by-step explanation:

Just type in a calc

Answer:

[tex]\sqrt{11}=3.31[/tex], -2.5, [tex]2\frac{1}{2}= 2.25[/tex], 3.6, -3.97621...

Step-by-step explanation:

Greatest to least would be:

3.6,  [tex]\sqrt{11}[/tex], [tex]2\frac{1}{4}[/tex], -2.5, -3.97621...

Least to greatest would be:

-3.97621, -2.5, [tex]2\frac{1}{4}[/tex], [tex]\sqrt{11}[/tex], 3.6

Hopefully, that helps.

Suppose that 5 people should be randomly selected from a group of 20 forming couples by 10. What is the probability that the 5 unrelated chosen from related persons (that is, no chosen person be a couple)?

Answers

The probability that none of the 5 randomly selected individuals are part of a couple is 0.016.

What is the probability that none of the 5 randomly selected individuals are part of a couple?

A probability means the branch of math which deals with finding out the likelihood of the occurrence of an event. Its measures the chance of an event happening.

We will know total number of possible outcomes when selecting 5 individuals from a group of 20. This can be calculated using the combination formula:

C(20, 5) = 20! / (5! * (20 - 5)!)

C(20, 5) = 15,504

We know that when we select an individual, we are removing their corresponding partner from the pool of available choices. This means that for each individual we choose, the number of available choices decreases by 1.

The number of favorable outcomes can be calculated as follows:

= 20 * 18 * 16 * 14 * 12

= 967,680

The probability will be:

= Outcomes / Favorable outcomes

= 15,504 / 967,680

= 0.01602182539

= 0.016.

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2.1. let a be the event that 2 consecutive flips both yield heads and let b be the event that the first or last flip yields tails. prove or disprove that events a and b are independent.

Answers

The events A and B are not independent. The occurrence of event B affects the probability of event A.

To determine whether events A and B are independent,

we need to check if the probability of event A occurring is affected by the occurrence of event B, and vice versa.

Probability of event A: Since we are flipping two coins,

the probability of getting heads on each flip is 1/2.

Therefore, the probability of getting two consecutive heads is

[tex](1/2) \times (1/2) = 1/4[/tex]

Probability of event B: The first or last flip yielding tails means there are two possibilities:

either the first flip is tails and the second flip is any outcome,

or the first flip is any outcome and the second flip is tails.

Each of these individual possibilities has a probability of

[tex](1/2) \times (1/2) = 1/4[/tex]

Hence, theprobability of event B is 1/4 + 1/4 = 1/2.

Since the probability of event A is 1/4 and the probability of event B is 1/2, and 1/4 ≠ 1/2,

we can conclude that events A and B are not independent.

The occurrence of event B (first or last flip yielding tails) affects the probability of event A (two consecutive flips yielding heads).

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[tex]\left \{ {{x=3} \atop {y+1=0}} \right.[/tex] solve graphically this linear system of equations

Answers

Answer:

The solution is the point (3, -1)

Step-by-step explanation:

We have the system of equations:

x = 3

y + 1 = 0

To solve this graphically, we need to graph these two lines and see in which point the lines intersect.

To graph the line x = 3, we need to draw a vertical line that passes through x = 3.

To graph y + 1 = 0

First we should isolate y.

y = -1

This is graphed as a horizontal line that passes through y = -1

The graph of these two lines can be seen in the image below.

Where the green line is x = 3, and the blue line is y = -1

Now, looking at the graph we can see that the lines do intersect in the point (3, -1)

Then the solution of the system is the point (3, -1)

What correction does Petri need to make? BRAINLIEST I WILL CROWN

Answers

You need to subtract 5-3 first because of PEMDAS petri did it wrong because she did 5 to the power of 2 and 2 to the power of 2 instead of actually subtracting 5-3. So the answer is the last option:)

Please help me I need to answer this question

Answers

Answer:

.........

it is -7

IM GIVING BRAINLIEST!!PLEASE HELP!!

Answers

Answer: c

Step-by-step explanation:

A medical team randomly selects people in an area, until he finds a person who has a corona virus, Let p is the probability that he succeeds in finding such a person, is 0.2 and X denote the number of people asked until the first success. (i) What is the probability that the team must select 4 people until he finds one who has a corona virus? (ii) What is the probability that the team must select more than 6 people before finding one who who has a corona virus?

Answers

Answer : i) The probability of finding the first case in 4 trials is 0.1024                        ii) The probability that the team must select more than 6 people before finding one who has a corona virus is 0.4095.

Explanation : Given information:Let p is the probability that he succeeds in finding such a person, is 0.2 and X denote the number of people asked until the first success.

(i) What is the probability that the team must select 4 people until he finds one who has a corona virus?  

The number of trials required until the first success follows geometric distribution.          

The probability of finding the first case in 4 trials is: P(X = 4) = q^3p, where q = 1 - p.                                                                                                                                     We have p = 0.2 and q = 0.8. So, P(X = 4) = 0.8^3 × 0.2 = 0.1024

(ii) What is the probability that the team must select more than 6 people before finding one who who has a corona virus?                                                                                                                                                                        P(X > 6) = 1 - P(X ≤ 6)                                                                                          The probability of finding the first case in the first 6 trials is:P(X ≤ 6) = 1 - q^6p= 1 - 0.8^6 × 0.2= 0.59049P(X > 6) = 1 - P(X ≤ 6)= 1 - 0.59049= 0.4095                                                                                                                Therefore, the probability that the team must select more than 6 people before finding one who has a corona virus is 0.4095.

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