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3 years ago

(GIVING BRAINLIEST!!)

Mathematics

2 answers:

Kipish [7]3 years ago

8 0

Answer:

b,c

Step-by-step explanation:

stepladder [879]3 years ago

6 0

Answer: B

Step-by-step explanation:

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At an archery competition, Aylin hits the target 14 out of 20 tries in round one. If Aylin gets 24 tries in round two, approxima

Andre45 [30]

Answer:

16.8

Step-by-step explanation:

14/20 = 0.7

24 x 0.7 = 16.8

(Feel free to round up or down, or keep it the same depending on the criteria of the question!)

6 0

3 years ago

A ski rental store rents a pair of skis for $44 a day and a snowboard for $58 a day. Yesterday, the store made $2,232 on ski and

Katen [24]

First, let the number of skis rented by x and the number of snowboards rented by y. We can then assemble the first equation from the amount of money made from the rentals.

44x + 58y = 2232

The second equation can come from the fact that 9 more skis were rented than snowboards.

y = x - 9

Therefore our system is:

44x + 58y = 2232

y = x - 9

3 0

4 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

3 years ago

PLz help me this is 7th grade math

UkoKoshka [18]

Answer:

g = 118°, h = 62°, k = 140°, m = 40°

Step-by-step explanation:

g = 118°

h = 180 - 118 = 62°

k = 140°

m = 180 - 140 = 40°

5 0

3 years ago

2800 to 2100 percent of change

Gennadij [26K]

2800-2100=700
2800/700=4
4% decrease

7 0

3 years ago