All categories

3 years ago

Find the area of a circle that has a diameter = 18

Mathematics

1 answer:

ki77a [65]3 years ago

4 0

Answer:

About 254.469 square units

Step-by-step explanation:

To find the area of a circle, you need to multiply the square of the radius by pi. To find the radius of a circle, you need to divide the diameter by 2, which in this case is 18/2=9. Now, you can plug it into the formula to find that the area of this circle is $\pi \cdot 9^2 \approx 254.469$ . Hope this helps!

You might be interested in

What is the equivalent fraction for 3,2,6,4 whole number?

maw [93]

3/1, 2/, 6/, 4/1, hope I helped!

8 0

3 years ago

A company wants to decrease their energy use by 16%.if their electric bill currently 2,2000 a month what will their be if they a

lyudmila [28]

The bill will be 18,480 if they are successful.

Step-by-step explanation:

Given,

Current bill of company = 22000

Energy usage to decrease = 16%

Amount of decreased usage = 16% of current bill

Amount of decreased usage = $\frac{16}{100}*22000$

Amount of decreased usage = $0.16*22000$

Amount of decreased usage = 3520

Bill after decreased usage = Current bill - Amount of decreased usage

Bill after decreased usage = 22000 - 3520 = 18480

The bill will be 18,480 if they are successful.

Keywords: percentage, subtraction

Learn more about subtraction at:

#LearnwithBrainly

8 0

3 years ago

The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the co

statuscvo [17]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if rising the process temperature reduces the thickness of the plastic film that coats a substrate material To do so, two samples of substrates are coated at different temperatures:

Sample 1

X₁: Thickness of the plastic film after the substrate is coated at 125F

n₁=11

X[bar]₁= 101.28

S₁= 5.08

Sample 2

X₂: Thickness of the plastic film after the substrate is coated at 150F

n₂= 13

X[bar]₂= 101.70

S₂= 20.15

Does the data support this claim? Use the P-value approach and assume that the two population standard deviations are not equal.

Now if the higher the heat, the thinner the thickness of the plastic coating, then the average thickness of the coating done at 150F should be less than the average thickness of the coating done at 125F, symbolically: μ₂ < μ₁

Then the hypotheses are:

H₀: μ₂ ≥ μ₁

H₁: μ₂ < μ₁

α:0.05 (there is no α level stated so I've chosen the most common one)

Assuming that both variables have a normal distribution since the population standard deviations are not equal, the statistic to use is the Welch's t-test:

$t= \frac{(X[bar]_2-X[bar]_1)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } } ~~t_w$

$t_{H_0}= \frac{(101.7-101.28)-0}{\sqrt{\frac{(20.15)^2}{13} +\frac{(5.08)^2}{11} } } = 0.072$

This test is one-tailed to the left, meaning that you'll reject the null hypothesis at small values of t. The p-value is also one-tailed and has the same direction as the test. To calculate it you have to first calculate the degrees of freedom of the Welch's t:

$Df_w= \frac{(\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S^2_1}{n_1})^2 }{n_1-1}+\frac{(\frac{S^2_2}{n_2} )^2}{n_2-1} }$

$Df_w= \frac{(\frac{5.08^2}{11} +\frac{20.15^2}{13} )^2}{\frac{(\frac{5.08^2}{11})^2 }{10} +\frac{(\frac{20.15^2}{13} )^2}{12} } = 13.78$

The distribution is a Student's t with 13 degrees of freedom, then you can calculate the p-value as:

P(t₁₃≤0.072)= 0.4718

Using the p-value approach, the decision rule is:

If the p-value ≤ α, the decision is to reject the null hypothesis.

If the p-value > α, the decision is to not reject the null hypothesis.

The p-value is greater than the significance level, so the decision is to nor reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to support the claim that the average thickness pf the plastic coat processed at 150F is less than the average thickness pf the plastic coat processed at 125F.

I hope you have a SUPER day!

4 0

3 years ago

QUESTION 4 If 1 foot = 0.37 metres and 1 mile = 5,180 feet, find the number of kilometres in 15 miles. show how to work out the

Andreas93 [3]

Answer:

28.749 km

Step-by-step explanation:

in reality

1 foot = 0.3048 meters not 0.37

1 mile = 5,280 feet not 5,180

15 miles = 24.14016 kilometers

---------------------------

Using the numbers stated in the problem

0.37m/ft * 5180ft/mile * 15mile * 1km/1000m = 28.749 km

4 0

2 years ago

Please answer my questions without links lol

guapka [62]

Answer:

Did you got my answer

if you don't then you can ask again as I will try my best

5 0

3 years ago