What is y? ( thank you for anyone who helps)

26% of college students live in campus housing. a random sample of 15 college students is selected. find the probability that 10

Sloan [31]

Answer:

The probability that 10 or more students will live in campus housing is approximately 0.111%

Step-by-step explanation:

Binomial distribution formula:

If the probability of success is $p$ , then the probability of $r$ successes out of $n$ trials will be: $^nC_{r}(p)^r(1-p)^n^-^r$

26% of college students live in campus housing. So here, $p= 26\%= 0.26$

A random sample of 15 college students is selected. So, $n= 15$

Now, "10 or more students" means 10 or 11 or 12 or 13 or 14 or 15 students. That means, $r= 10, 11, 12, 13, 14, 15$

Thus, the probability that 10 or more students will live in campus housing will be........

$[^1^5C_{10}(0.26)^1^0(1-0.26)^5]+[^1^5C_{11}(0.26)^1^1(1-0.26)^4]+[^1^5C_{12}(0.26)^1^2(1-0.26)^3]+[^1^5C_{13}(0.26)^1^3(1-0.26)^2]+[^1^5C_{14}(0.26)^1^4(1-0.26)^1]+[^1^5C_{15}(0.26)^1^5(1-0.26)^0]\\ \\ = 0.00094...+0.00015...+0.00001...+ 0.0000014...+ 0.000000071...+0.0000000016...\\ \\ =0.00111002... =0.111002...\% \approx 0.111\%$

3 0

3 years ago

Given ADC ACB and BDC BCA prove a squared + b squared = c squared. Use the two Column proof.

Yuri [45]

Answer:

a² + b² = c · (e + d) = c × c = c²

a² + b² = c²

Please see attachment

Step-by-step explanation:

Statement, Reason

ΔADC ~ ΔACB, Given

AC/AD = BA/AC, The ratio of corresponding sides of similar triangles

b/e = c/b

b² = c·e

ΔBDC ~ ΔBCA, Given

BC/BA = BD/BC, The ratio of corresponding sides of similar triangles

a/c = d/a

a² = c·d

a² + b² = c·e + c·d

a² + b² = c · (e + d)

e + d = c, Addition of segment

a² + b² = c × c = c²

Therefore, a² + b² = c²

7 0

3 years ago

1) A traditional die is a cube with each of its six sides representing the numbers from 1 to 6. Mr. Dicey customized a die by re

liraira [26]

Answer:

1) $\frac{2}{5}$

2) 49.225

3) $\frac{7}{2}$

Step-by-step explanation:

1) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1/1, 1/2, 1/3, 1/4, 1/5 and 1/6 and the probavility for all values is 1/6 so the expected values will be: $E(x)=(\frac{1}{1} *\frac{1}{6}) +(\frac{1}{2} *\frac{1}{6}) +(\frac{1}{3} *\frac{1}{6})+(\frac{1}{4} *\frac{1}{6})+(\frac{1}{5} *\frac{1}{6})+(\frac{1}{6} *\frac{1}{6})$

$E(x)=0.167+0.083+0.056+0.042+0.033+0.028=0.409\approx \frac{2}{5}$

2) To find the variance of the expected values we can use the equation:

$Var(x)=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2} }{n}$

So for our problem will be:

$Var(x)=\frac{(3-10.5)^2+(4-10.5)^2+(17-10.5)^2+(18-10.5)^2}{4}$

$Var(x)=\frac{56.25+42.25+42.25+56.25}{4}$ $Var(x)=\frac{196.9}{4}=49.225$

3) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1, 2, 3, 4, 5 and 6 and the probavility for all values is 1/6 so the expected values will be: $E(x)=(1*\frac{1}{6}) +(2 *\frac{1}{6}) +(3 *\frac{1}{6})+(4 *\frac{1}{6})+(5 *\frac{1}{6})+(6 *\frac{1}{6})$