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4 years ago

10

WILL MARK BRAINLIEST Solve the following problems. Write the complete proof in your paper homework and for online (only) complet

e the probing statement (if any) that is a part of your proof or related to it. b Given: FD ⊥ AB ∠A ≅ ∠B Prove: ∠F ≅∠CEF

1 answer:

RideAnS [48]4 years ago

5 0

Answer:

AED congruent to FEC

Step-by-step explanation:

If you look closely, BOth of the angles are vertical.

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xeze [42]

Answer:

-4

Step-by-step explanation:

So we are given the expression: | y - x | + y - 1, and we have our values for (x,y): (-3,-6).

So, we can just start plugging things in and simplifying, so:

| -6 + 3 | - 6 - 1 = | -3 | -7

= 3 - 7

= -4

7 0

3 years ago

Factor 6a2 - 10a - 24.

finlep [7]

The answer of the question is A

3 0

3 years ago

The Ohio Department of Agriculture tested 203 fuel samples across the state

Rus_ich [418]

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The proportion estimated would be:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

3 0

3 years ago

Read 2 more answers

If f(x) = 4x + 21 and g(x) = 2x + 2 what is f[g(7)]

Nana76 [90]

Answer:

we conclude:

$f(g(7)) =85$

Step-by-step explanation:

Given

$f(x) = 4x + 21$

$g(x) = 2x + 2$

To determine

$f(g(7)) =?$

In order to determine f(g(7)), the first step we need to do is to determine g(7)

so substitute x = 7 in g(x) = 2x + 2

$g(x) = 2x + 2$

$g(7) = 2(7) + 2$

$g(7) = 14 + 2$

$g(7) = 16$

so now we have

$f(g(7)) =f(16)$

The next step to do is to substitute x = 16 in f(x) = 4x + 21

$f(x) = 4x + 21$

$f(16) = 4(16) + 21$

$f(16) = 64 + 21$

$f(16) = 85$

so we have

$f(g(7)) =f(16)=85$

Therefore, we conclude:

$f(g(7)) =85$

6 0

3 years ago

Read 2 more answers

Jessie wants to buy three tickets for a concert. she has earned $135 and each tickey costs $50. she borrows the rest of the mone

dybincka [34]

Answer: In explanation

Step-by-step explanation:

1. 135 - 50x

2. $15

3. $0

I hope this helped!

3 0

1 year ago