Help me please!!!
10pts

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer:

No solutions.

Step-by-step explanation:

9|x-1|=-45

Divide 9 into both sides.

|x-1| = -45/9

|x-1| = -5

Absolute value cannot be less than 0.

Answer:

No solution

Step-by-step explanation:

=> 9|x-1| = -45

Dividing both sides by 9

=> |x-1| = -5

Since, this is less than zero, hence the equation has no solutions

Answer:

[tex]z=-\frac{20}{3}[/tex]

Step-by-step explanation:

[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]

Best Regards!

Answer:

the answer is 3

Step-by-step explanation:

i took the test

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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Answer/Step-by-step explanation:

When solving problems like this, remember the following:

1. + × + = +

2. + × - = -

3. - × + = -

4. - × - = +

Let's solve:

a. (-4) + (+10) + (+4) + (-2)

Open the bracket

- 4 + 10 + 4 - 2

= - 4 - 2 + 10 + 4

= - 6 + 14 = 8

b. (+5) + (-8) + (+3) + (-7)

= + 5 - 8 + 3 - 7

= 5 + 3 - 8 - 7

= 8 - 15

= - 7

c. (-19) + (+14) + (+21) + (-23)

= - 19 + 14 + 21 - 23

= - 19 - 23 + 14 + 21

= - 42 + 35

= - 7

d. (+5) - (-10) - (+4)

= + 5 + 10 - 4

= 15 - 4 = 11

e. (-3) - (-3) - (-3)

= - 3 + 3 + 3

= - 3 + 9

= 6

f. (+26) - (-32) - (+15) - (-8)

= 26 + 32 - 15 + 8

= 26 + 32 + 8 - 15

= 66 - 15

= 51

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Answer:

Question 1: 3 - 5 hours.

Question 2: 0 - 1 hour

Step-by-step explanation:

Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5  hours.

Question 2:  In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.

Answer:

20gallons

Step-by-step explanation:

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

(a)The total sales of ice-cream last month is 702 gallons.

(b)The total sales of broccoli last month is 510 pounds.

Step-by-step explanation:

Part A

Total Sales of gallons of ice cream this month = 597

Since it represents a decrease of 15% of last month's sales

Let the total sales of ice-cream last month =x

Then:

(100-15)% of x =597

85% of x=597

0.85x=597

x=597/0.85

x=702 (to the nearest integer)

The total sales of ice-cream last month is 702 gallons.

Part B

Total Sales of broccoli this month = 617 pounds

Since it represents an increase of 21% of last month's sales

Let the total sales of ice-cream last month =y

Then:

(100+21)% of y =617

121% of y=617

1.21y=617

y=617/1.21

y=510 (to the nearest integer)

The total sales of broccoli last month is 510 pounds.

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

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Answer:

a) Probability that a team will win the match given that it has won the first game = 0.66

b) Probability that a team will win the match given that it has won the first two games= 0.81

c) Probability that a team will win the match, given that it has won two out of the first three games = 0.69

Step-by-step explanation:

There are a total of seven games to be played. Therefore, W and L consists of 2⁷ equi-probable sample points

a) Since one game has already been won by the team, there are 2⁶ = 64 sample points left. If the team wins three or more matches, it has won.

Number of ways of winning the three or more matches left = [tex]6C3 + 6C4 + 6C5 + 6C6[/tex]

= 20 + 15 + 6 + 1 = 42

P( a team will win the match given that it has won the first game) = 42/64 = 0.66

b)  Since two games have already been won by the team, there are 2⁵ = 32 sample points left. If the team wins two or more matches, it has won.

Number of ways of winning the three or more matches left = [tex]5C2 + 5C3 + 5C4 + 5C5[/tex] = 10 + 10 + 5 +1 = 26

P( a team will win the match given that it has won the first two games) = 26/32 = 0.81

c) Probability that a team will win the match, given that it has won two out of the first three games

They have played 3 games out of 7, this means that there are 4 more games to play. The sample points remain 2⁴ = 16

They have won 2 games already, it means they have two or more games to win.

Number of ways of winning the three or more matches left = [tex]4C2 + 4C3 + 4C4[/tex] = 6 + 4 + 1 = 11

Probability that a team will win the match, given that it has won two out of the first three games = 11/16

Probability that a team will win the match, given that it has won two out of the first three games = 0.69

Answer:

  (x, y) = (7, 4) meters

Step-by-step explanation:

The area of the floor without the removal is x^2, so with the smaller square removed, it is x^2 -y^2.

The perimeter of the floor is the sum of all side lengths, so is 4x +2y.

The given dimensions tell us ...

  x^2 -y^2 = 33

  4x +2y = 36

From the latter equation, we can write an expression for y:

  y = 18 -2x

Substituting this into the first equation gives ...

  x^2 -(18 -2x)^2 = 33

  x^2 -(324 -72x +4x^2) = 33

  3x^2 -72x + 357 = 0 . . . . write in standard form

  3(x -7)(x -17) = 0 . . . . . factor

Solutions to this equation are x=7 and x=17. However, for y > 0, we must have x < 9.

  y = 18 -2(7) = 4

The floor dimension x is 7 meters; the inset dimension y is 4 meters.

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

Just replace the variables with the number

d5

c4 (uh oh)

a2

b-3

f-7

d-c = 5 - 4 = 1

1/3 - 4(ab+f)

2 x -3 = -6

-6 + -7 = -13

-13 x 4 = -52

1/3 - -52 = 1/3 + 52 =

52 1/3

Hope this helps

Step-by-step explanation:

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

25%

Step-by-step explanation:

The last percentile always contains 25% of the observations.

Answer:

I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]

Step-by-step explanation:

Answer:

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n=900[/tex] represent the sample size selected

[tex]p = 0.75[/tex] represent the population proportion

We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:

[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]

And the standard error is given;

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]

Answer:

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:  b) the tiles are not similar because both SP:SR is 5:4 and MJ:ML is 5:2

Step-by-step explanation:

We are given that the tiles are rectangular which implies that they both have a 90° angle.

In order to prove similarity, We need to show that the lengths and widths are proportional.

P Q R S

J  K L M

a) PQ : QR         JK : LM

  w=4  L=5        w=2 w=2

                                  ↓

                                 We need Length (not width)

b) SP : SR         MJ : ML

  L=5  w=4        L=5 w=2

      5 : 4               5 : 2

When comparing length to width they do not have the same ratio so the rectangles are not similar.

c) PQ : QR         JK : KL

  w=4  L=5        w=2 L=5

      4 : 5               2 : 5

When comparing width to length they do not have the same ratio so the rectangles are not similar.

d) SR : ML         PQ : JK

  w=4  w=2        w=4 w=2

           ↓                     ↓

  We need Length (not width)

Answer:

Height of tomato plant is the dependent variable

Presence of walnut leaves in the soil is the independent variable

Tomato plants grown without walnut leaves is the control

Step-by-step explanation:

An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.

A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.

A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.

Answer:

Is possible to make a Type I error, where we reject a true null hypothesis.

Step-by-step explanation:

We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.

As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.

The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.

This is a Type I error, where we reject a true null hypothesis.

Answer: 3.9

Step-by-step explanation: Khan Academy

The length of BD if The angle ∠ DAC is equal to the angle ∠ BAD is 3.92.

Three straight lines coming together create a triangle. There are three sides and three corners on every triangle (angles). A triangle's vertex is the intersection of two of its sides. Any one of a triangle's three sides can serve as its base, however typically the bottom side is used.

Given:

The angle ∠ DAC = angle ∠ BAD

As we can see that the triangle BAD and triangle DAC are similar By SAS similarity,

AC / AB = CD / BD  (By the proportional theorem of similarity)

5.6 / 5.1 = 4.3 / BD

1.09 = 4.3 / BD

BD = 4.3 / 1.09

BD = 3.92

Thus, the length of BD is 3.92.

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