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

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

Read 2 more answers

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