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mixas84 [53]
3 years ago
14

Amy ran 3/4 of a mile. Convert into decimal. Show work.

Mathematics
2 answers:
mezya [45]3 years ago
5 0
3/4 as a decimal is 0.75
kicyunya [14]3 years ago
4 0
3/4 of 100 is 75 so 3 quarters of a mile is 0.75
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leah wrote two different fractions with the same denominator.both fractions were less than 1.can their sum equal 1? can their su
dem82 [27]
Their sum can be both equal, or greater. If you have 3/8 and 6/8, you would get 9/8 or 1 and 1/8. Which is greater than 1. Also if you add 2/6 and 4/6, you get 6/6 or 1. proving that both are possible.
3 0
3 years ago
In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal
ddd [48]

Answer:

a) z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236  

p_v =2*P(Z>0.236)=0.813  

If we compare the p value and using any significance level for example \alpha=0.01 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.  

b) We are confident at 99% that the difference between the two proportions is between -0.188 \leq p_B -p_A \leq 0.226

Step-by-step explanation:

Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_A =\frac{26}{75}=0.347 represent the estimated proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_A=75 is the sample size for A

p_B represent the real population proportion for women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_B =\frac{21}{64}=0.328 represent the estimated proportion of women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_B=64 is the sample size required for B

z represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:  

Null hypothesis:p_{A} = p_{B}  

Alternative hypothesis:p_{A} \neq p_{B}  

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

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

Where \hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338

Calculate the statistic

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

z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236  

Statistical decision

Since is a two sided test the p value would be:  

p_v =2*P(Z>0.236)=0.813  

If we compare the p value and using any significance level for example \alpha=0.01 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.  

Part b  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188  

(0.347-0.328) + 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=0.226  

And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).  

We are confident at 99% that the difference between the two proportions is between -0.188 \leq p_B -p_A \leq 0.226

5 0
3 years ago
Help quickly as possible thank you :) ​
Stells [14]

Answer:

50/100 and .50

Step-by-step explanation:

6 0
2 years ago
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A bag contains 11 purple, 19 orange, and 12 blue tennis balls. A person picks out an orange tennis
AlexFokin [52]

The answer is Yes, Hope this helps !

7 0
3 years ago
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Plz help me i dont even know how to start this can u also explain how to do it much appretciated
Sergeu [11.5K]
Ok did your teacher say anything about not being able to use negative numbers or fractions? If not can you use fractions or negative numbers.
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