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3 years ago

To the nearest tenth, what is the area of the shaded segment when JA = 8 ft?

1 answer:

3 0

I think you forgot to give the diagram along with the question. Considering 8 ft as the radius, we can find the area by the formula described below.
Area = pi * (Radius) ^2
= 3.14 * (8)^2
= 3.14 * 64
= 200.96
So the area of the shaded portion would be 200.96 square feet. I hope that this answer has come to your help.

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Solve for y=7x-3w for x

il63 [147K]

You would get that X equals 3w over 7 + y over 7<span />

5 0

3 years ago

I need help with 16a and 17. i haven't been able to understand this in class and no one was able to help me

kompoz [17]

16a. 21cm. bc when it has 10 grams it moves half of that to 5cm so divide 42 by 2 and you get 21cm

17. remember y=mx+b; m is slope and b is y-intercept
you have to put your equation from point slope to y=mx+b
4x+y=6
subtract 4x from each side
y=-4x+6
slope = -4
y-intercept= 6

hope this helped :)

5 0

3 years ago

Levi borrowed $7129 at 4% simple interest for 6 months. How much will the interest amount to? What is the total amount that Levi

rodikova [14]

Answer:

Interest will amount to $142.58

Total amount Levi will have to pay back at the end of 6 months=$7271.58

Step-by-step explanation:

Simple interest=principal×rate×time

Principal=$7129

Rate=4%=0.04

Time=6 month=0.5 year

Simple interest=principal×rate×time

=$7129×0.04×0.5

=$142.58

Interest will amount to $142.58

Total amount Levi will have to pay back at the end of 6 months

=principal + total interest

=$7129+$142.58

=$7271.58

4 0

3 years ago

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 20

Murljashka [212]

Answer:

There is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1432

p = 23% = 0.23

Alpha, α = 0.05

Number of theft complaints , x = 321

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.23\\H_A: p < 0.23$

This is a one-tailed test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{321}{1432} =0.2241$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.2241-0.23}{\sqrt{\frac{0.23(1-0.23)}{1432}}} = -0.5305$

Now, we calculate the p-value from the table.

P-value = 0.298

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.

5 0

3 years ago

A fair number cube with

malfutka [58]

Answer:

b=15 percentage

a=20 percentage

6 0

3 years ago