Taking a test!!! I need help on it pleaseee!!!

involving a carnival fishpond. Assume that the pond contains 100 fish: 78 purple, 21 blue, and 1 silver. A contestant pays $0.65

saw5 [17]

Answer:

On average the carnival gain on each play

= 0.04 dollars

Step-by-step explanation:

Given that a pond contains 100 fish: 78 purple, 21 blue, and 1 silver.

Fish Purple Blue Silver total

Frequency 78 21 1 100

Prob 0.78 0.21 0.01 1

Revenue 0.4 0.8 13

game fee 0.65 0.65 0.65

Net revenue -0.25 0.15 12.35

Net Rev*Prob -0.195 0.0315 0.1235 -0.04

Thus we get per player expected net revenue is -0.04

This would be gain for Carnival

Hence On average the carnival gain on each play

= 0.04 dollars

3 0

3 years ago

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts o

Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A $(100-\alpha )\%$ confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a $(100-\alpha )\%$ probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the t-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

$\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}$

Here,

$\bar x$ = sample mean

s = sample standard deviation

n = sample size

$t_{\alpha/2,(n-1)}$ = critical value.

The degrees of freedom for the critical value is, (n - 1) = 7 - 1 = 6.

The significance level is: $\alpha =1-Confidence\ level=1-0.99=0.01$

The critical value is:

$t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143$

**Use the t-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753$

$\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379$

The 99% confidence interval for μ is:

$x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)$

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

6 0

3 years ago

A city council has to use its resources to improve either the water supply or the drainage system. At which level is this decisi

Fynjy0 [20]

A city council has to use its resources to improve either the water supply or the drainage system. At which level is this decision made?

The asnswer is: Government

5 0

3 years ago

Which of the following numbers is a multiple of 6? A. 333 B. 882 C. 424 D. 106

gladu [14]

If a number is a multipule of 6 then it is a multiplule of 2 AND 3

if a number is a multiplule of 2 then the last digit should be divisible by 2 (ie, 0,2,4,6,8)

if a number is divisible by 3, then the sum of its digits is divisble by 3
try each

A. 333
last number is not divisible by 2 so not divisible by 6

B. 882
2 is divisible by 2
8+8+2=18
18/3=6
divisible by 6 yes

C. 424
4 is disibielbe by 2
4+2+4=10
10/3=not a whole number so not divisible by 6

D. 106
6 is divisible by 2
1+0+6=7
7/3= not a whole number so not divibislb eby 6

answer is B

6 0

3 years ago

Read 2 more answers

I want solution of ulike unsolved paper 2 Q.23

irakobra [83]

We need the question to help

5 0

3 years ago