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2 years ago

hellohello, this needs to be turned in tomorrow so if you could help, i would really appreciate it. :]

Mathematics

1 answer:

Marizza181 [45]2 years ago

5 0

Answer:

24 is the answer

hope this helps

have a good day :)

Step-by-step explanation:

You might be interested in

Your favorite video game just went on sale for 33% off.

abruzzese [7]

Answer:

$40.20

Step-by-step explanation:

60 / 100 = 0.6

0.6 x 33 = 29.8

60 - 29.8 = 40.2

Thus he price after the discount would be $40.20

3 0

2 years ago

(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

7 0

2 years ago

Please help :) im stupid and bad at this lol

My name is Ann [436]

Answer:

Addition. “Also, I have to stop at the store on the way home.” ...

Comparison. “In the same way, the author foreshadows a conflict between two minor characters.” ...

Concession. “Granted, you did not ask ahead of time.” ...

Step-by-step explanation:

4 0

2 years ago

What decimal is equal 35%

iren [92.7K]

Answer:

0.35

Step-by-step explanation:

Divide 35 by 100 and you get 0.35

8 0

3 years ago

Read 2 more answers

Help please Question is in the picture

anzhelika [568]

Answer:

360 degrees in total.

Step-by-step explanation:

We are given that the m∠1 is 145 degrees, and because ∠1 and ∠2 form a straight line we can tell that 180-145 = 35.

So now here's what we know:

m∠1 = 145

m∠2 = 35

After that, we need to look at ∠5 and ∠6.

One thing we can notice right away is that ∠5 and ∠6 should measure the same as ∠1 and ∠2 (respectively) because they are corresponding angles.

Now we will add them all together to find the sum of all four angles.

m∠1 = 145 + m∠2 = 35 = 180

m∠5= 145 + m∠6 = 35 = 180

180 + 180 = 360

8 0

3 years ago