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sp2606 [1]
3 years ago
9

3.) What is the diameter and circumference of a circle given the area is 16

Mathematics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

D =4.51 and the

Area is=16

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PLEASE HELP (50 Points)....
Vedmedyk [2.9K]

Total deposit = 645.30 + 645.30 = $1290.60

Total withdrawals= 158.25 +168.36+157.42+148.46 = $632.49

Balance = 398.26 + 1290.60 - 632.49 = $1056.37

4 0
3 years ago
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1.36 km is how many cm
Taya2010 [7]

Answer:

136000 cm

Step-by-step explanation:

I think it's helps you

5 0
3 years ago
Solve the following inequality for dd. Write your answer in simplest form. −7d−3>2d-10
LUCKY_DIMON [66]

Answer:

  d < 7/9

Step-by-step explanation:

Add 7d +10

  7 > 9d

Divide by 9

  7/9 > d

3 0
3 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
Atown uses positive numbers to track increases in population and negative numbers to track decreases in population.
amid [387]

Answer:

Step-by-step explanation:

-60

Step-by-step explanation:

Increase in population is given by positive sign

Change in population in 5 years = -300

Decrease in population is represented by negative sign

Change in population in 1 year = -300\5

                                                = -60

Town's population faces a decrease of 60 residents each year.

4 0
3 years ago
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