<img src="https://tex.z-dn.net/?f=7.75%20divided%20by%204%201%2F2" id="TexFormula1" title="7.75 divided by 4 1/2" alt="7.75 divi

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Vladimir [108]

2 years ago

ded by 4 1/2" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Andrew [12]2 years ago

5 0

Answer:

2.64516129

Step-by-step explanation:

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I need the answer for this problem please

Savatey [412]

Answer:

Step-by-step explanation:

Find all probabilities:

A. False

$Pr(\text{red shirt}|\text{large shirt})=\dfrac{\text{number red large shirts}}{\text{number large shirts}}=\dfrac{42}{77}=\dfrac{6}{11}\\ \\Pr(\text{large shirt})=\dfrac{\text{number large shirts}}{\text{number shirts}}=\dfrac{77}{165}=\dfrac{7}{15}$

B. True

$Pr(\text{blue shirt}|\text{large shirt})=\dfrac{\text{number blue large shirts}}{\text{number large shirts}}=\dfrac{35}{77}=\dfrac{5}{11}\\ \\Pr(\text{blue shirt})=\dfrac{\text{number blue shirts}}{\text{number shirts}}=\dfrac{75}{165}=\dfrac{5}{11}$

C. False

$Pr(\text{shirt is medium and blue})=\dfrac{\text{number medium and blue shirts}}{\text{number shirts}}=\dfrac{48}{165}=\dfrac{16}{55}\\ \\Pr(\text{medium shirt})=\dfrac{\text{number medium shirts}}{\text{number shirts}}=\dfrac{88}{165}=\dfrac{8}{15}$

D. False

$Pr(\text{large shirt}|\text{red shirt})=\dfrac{\text{number red large shirts}}{\text{number red shirts}}=\dfrac{42}{90}=\dfrac{7}{15}\\ \\Pr(\text{red shirt})=\dfrac{\text{number red shirts}}{\text{number shirts}}=\dfrac{90}{165}=\dfrac{6}{11}$

4 0

2 years ago

In monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 6 day peri

liubo4ka [24]

Answer:

The margin of error that corresponds to a 95% confidence interval is 4.96.

Step-by-step explanation:

We have the standard error(which is the same as the standard deviation of the sample), so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. There are 6 days, so

df = 6 - 1 = 5

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{0.975}$ ). So we have T = 2.5706