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3 years ago

Can you help me plz.

Mathematics

2 answers:

adelina 88 [10]3 years ago

4 0

Answer:1.885 miles

Step-by-step explanation:Do 0.58 times 3.25 which gives you 1.885.

Hope it helps :)

Alexxandr [17]3 years ago

4 0

Answer:

1.885

Step-by-step explanation:

Mutiply 0.58 which is the miles per lap by 3.25 which are the number of laps Jessica walked

0.58*3.25=1.885

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WHERE DO I DRAG THE LAST TWO

Anastasy [175]

-3/4 goes between -1 and 0 it is the 3rd line to the left and n would go between positive 0 and 1 on the 3rd line to the right

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3 years ago

CS Algebra

CaHeK987 [17]

Answer:

Hope this helps :)

Step-by-step explanation:

8(x - 2) = 2x + 8

y+9 = -2(y + 1)

value of x in 8(x - 2) = 2x + 8

x=4

substitue

y+9=−2(y+1)

value of y y+9=−2(y+1)

y= - 11/3 or 3.66...

x=4

y=4 (I rounded 3.66)

4 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

HELP ASAP!!! I SUCK AT MATH AND I DON’T understand this!!!

anyanavicka [17]

Answer:

Part A:

$\frac{13}{20}$