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Norma-Jean [14]
3 years ago
7

A track coach made a two way table to show his runners gender and running category. Male: Sprints- 16 Middle distance- 11 Long d

istance- 9 Female: Sprints- 12 Middle distance- 14 Long distance- 5 To the nearest hundredth, what is the probability that a randomly chosen long distance runner is male?
Mathematics
2 answers:
sasho [114]3 years ago
8 0

Answer: Our required probability is 0.134.

Step-by-step explanation:

Since we have given that

                    Sprints           Middle distance        Long distance      Total

Male                 16                          11                               9                   36

Female              12                          14                              5                   31

Total                 28                         25                             14                   67

So, the probability that a random chosen long distance runner is male is given by

\dfrac{\text{Number of male for long distance}}{\text{Total outcomes}}\\\\=\dfrac{9}{67}\\\\=0.134

Hence, our required probability is 0.134.

hram777 [196]3 years ago
3 0
There are 9 long distance male runners and 5 long distance female runners. This makes a total of 9+5 = 14 long distance runners.

Out of this total of 14 long distance runners, there are 9 males. So the probability as a fraction is 9/14

If you need the answer in decimal form, then use a calculator to get 9/14 = 0.642857 which is approximate.
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Question 24 Multiple Choice Worth 1 points)
hammer [34]

Given:

The system of equations is:

Line A: y=x-4

Line B: y=3x+4

To find:

The solution of given system of equations.

Solution:

We have,

y=x-4              ...(i)

y=3x+4           ...(ii)

Equating (i) and (ii), we get

x-4=3x+4

-4-4=3x-x

-8=2x

Divide both sides by 2.

-4=x

Substituting x=-4 in (i), we get

y=-4-4

y=-8

The solution of system of equations is (-4,-8).

Now verify the solution by substituting x=-4, y=-8 in the given equations.

-8=-4-4

-8=-8

This statement is true.

Similarly,

-8=3(-4)+4

-8=-12+4

-8=-8

This statement is also true.

Therefore, (-4,-8) is a solution of the given system of equations, because the point satisfies both equations. Hence, the correct option is C.

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2 years ago
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Inga [223]
Try making a proportion 10,000/1 and then 1000/x and solve
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3 years ago
A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

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Answer:

2(2g+1)

Explanation:

4g and 2 goes into 2

divide by 2

4g/2=2g

2/2=1

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3 years ago
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I think its letter A
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3 years ago
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