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olya-2409 [2.1K]

2 years ago

11

A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeles

s persons who are veterans.

1 answer:

MA_775_DIABLO [31]2 years ago

4 0

Answer:

[0.184, 0.266]

Step-by-step explanation:

Given:

Number of survey n =280

Number of veterans = 63

Confidence interval = 90%

Computation:

Probability of veterans = 63/280

Probability of veterans =0.225

a=0.1

Z(0.05) = 1.645 (from distribution table)

Confidence interval = 90%

So,

p ± Z*√[p(1-p)/n]

0.225 ± 1.645√(0.225(1-0.225)/280)

[0.184, 0.266]

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The Art Club decided to sell t-shirts in order to pay for their field trip to a nearby museum. They started with 460 t-shirts an

blondinia [14]

Step by Step Answer:
10% of 460 = 46
46 x 9 =414
The Art Club sold 414 shirts.

4 0

2 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

2 years ago

Help with the problem in picture

Flauer [41]

Answer:

The slope of a line is a measure of its steepness. Mathematically, slope is calculated as "rise over run" (change in y divided by change in x).

Step-by-step explanation:

7 0

1 year ago

How much does 1 candy bar cost if 112 candy bars cost $78.00?

Alexxandr [17]

$0.70 per candy.
$78 divided by 112 gives $0.696, so by rounding off, you get 0.70

5 0

2 years ago

I need help with math here the photos

arsen [322]

Answer:

9-49

a= 2r+ 7

b= 2.5x + 16

c= 3r-2x-8.8

Step-by-step explanation:

4 0

2 years ago