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2 years ago

Can you please help me

Mathematics

1 answer:

Levart [38]2 years ago

5 0

thanks for your help I really think that it's three

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There are 10 balls in an urn, numbered from 1 to 10. If 5 balls are selected at random and their numbers are added, what is the

Kisachek [45]

Let $B_i$ denote the value on the $i$ -th drawn ball. We want to find the expectation of $S=B_1+B_2+B_3+B_4+B_5$ , which by linearity of expectation is

$E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i]$

(which is true regardless of whether the $X_i$ are independent!)

At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.

$P(X_i=x)=\begin{cases}\frac1{10}&\text{for }x\in\{1,2,\ldots,10\}\\0&\text{otherwise}\end{cases}$

and so

$E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5$

Then the expected value of the total is

$E[S]=5(5.5)=\boxed{27.5}$

8 0

3 years ago

Simplify. <br> <img src="https://tex.z-dn.net/?f=%5Csqrt%7B44%7D" id="TexFormula1" title="\sqrt{44}" alt="\sqrt{44}" align="absm

zvonat [6]

The answer is 6.6
hope this helps :)

3 0

3 years ago

1) Erica bought a car for $24,000. She had to add Pennsylvania's sales tax of 6%. The total price of the car is closest to?

Lina20 [59]

To find the car's total price, add the tax paid to the car's original price.

$32800 + $1968 = $34768

The total cost is $34,768

7 0

3 years ago

Read 2 more answers

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a

Vikki [24]

Answer:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

Step-by-step explanation:

Information given:

$n=131$ represent the sample size

$\hat p=0.81$ represent the estimated proportion

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.