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2 years ago

Tiny's Tattoo Parlor sold 389 tattoos this month. 43 of those tattoos were arm tattoos.

Mathematics

1 answer:

Alexxx [7]2 years ago

3 0

Answer:

346/389

Step-by-step explanation:

First determine how many were not arm tattoos

389-43

346

P( not arm) = not arm /total

= 346/389

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Order these numbers from least to greatest.<br><br> 8.88, 8.8, 7.3962, 8.808

Wewaii [24]

7.3962, 8.8, 8.808, 8.88

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2 years ago

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(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

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2 years ago

Describe the transformation of the graph of the parent function y= sqrt x+7 + 5

stellarik [79]

Answer:

The function moves 7 to the left and 5 units up

Step-by-step explanation:

y = (x + 7) + 5

I think that's the problem you put.

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2 years ago

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the age of biken is two times more than that of mousam. if the sum of their age is 30 years than find their difference age with

Pavlova-9 [17]

$b = 2m \\ b + m = 30 \\ m = 10 \\ b = 20$

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2 years ago

HURRY I NEED HELP QUICK!!!!!!!

juin [17]

Answer:

Step-by-step explanation:

If you notice, point 4 on the x-axis (horizontal) coincides with 140 on the y-axis.

If you multiply x * 35 = 140

x is four.

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2 years ago