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ikadub [295]
2 years ago
6

Urgent help me plzzz this is urgent i will give brainilest

Mathematics
1 answer:
scoray [572]2 years ago
4 0

Answer:

to blury i cant see

Step-by-step explanation:

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Question and answer choices are in the SS attached!!!! Pls help me! ty!
victus00 [196]

Answer:

B

Step-by-step explanation:

It's a right triangle, so the pythagorean theorem will be used a lot.

First, find the side length on the left.

25^2 = 7^2 + y

y = 24

Now, knowing that 7 - 3 = 4 (the smaller side length), I can use this theorem again to find x.

4^2 + 24^2 = x^2

x = \sqrt{592}

Just need to simplify it a little farther...

592 / 16 = 37, so B must be the answer.

3 0
2 years ago
$8 buys 40 ounces of ground turkey
3241004551 [841]

Answer:

idk

Step-by-step explanation:

5 0
3 years ago
Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has
Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

Z_{1- \alpha /2}= Z_{0.90} = 1.28

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= \frac{(3.26-3.04)}{2} = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

n=  (0.33 * \frac{1.28}{0.08})²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0
3 years ago
Best Buy purchases a laptop for $244. They then sell the laptop for $927.20. What is the percent increase in the price?
maria [59]

Answer:

sjbs

Step-by-step explanation:

ywyeyeyjejsjeueuheuehe

6 0
2 years ago
Help <br> me to solve this please
Leona [35]

Answer:

for this one I think it mostly 3

3 0
2 years ago
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