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olga_2 [115]
2 years ago
5

You survey your 22 classmates on there favorite color. There are 396 students at your school. How many students in the school do

you predict would choose green as there favorite color
Mathematics
1 answer:
lana66690 [7]2 years ago
3 0

Answer:

The answer is 18 times the number given in the survey.

Step-by-step explanation:

The survey sample is an example of the 396 students. If there are 22 classmates in the survey, you need to turn 22 into 396 for it to work. 22 times 18 equals 396, therefore 18 times the number of green in the survey.

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Find the Volume of a cube with edges 2.2 millimeters long.
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Answer:

V≈10.65

a Edge=2.2

Solution

V=a3=2.23

≈10.648

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How many kilograms are there in 37 pounds?
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Answer:16.8

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A total of 505 tickets were sold for a school play they were either adult tickets or student tickets they were 55 more student t
arsen [322]

Hello There!

<u><em>The number of adult tickets sold in the school play = 225</em></u>

<u><em></em></u>

<u><em>Total Number Of Tickets = 505</em></u>

<u><em>Number Of Student Tickets = x + 55</em></u>

<u><em></em></u>

Number of adult tickets sold can be represented by X

Number of student tickets = X + 55

X+X+55=505

2x+55=505

2x=505-55

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7 0
3 years ago
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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!

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3 years ago
Neeeeedd help hurry please
Alona [7]

Answer:

Step-by-step explanation:

Sslsl

6 0
2 years ago
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