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Ivahew [28]
3 years ago
10

Use the grouping method to factor this polynomial completely 3x^3+12x^2+2x+8

Mathematics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

(x+4)(3x^2 + 2)

Step-by-step explanation:

3x^3+12x^2+2x+8

3x^2(x + 4) + 2(x + 4)

(x+4)(3x^2 + 2)

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Savannah has an action figure collection. She keeps 57 of the action figures on her wall, which is 19% of her entire collection.
beks73 [17]

Answer:

She has a total of 300 action figures.

Step-by-step explanation:

19 : 100 = 57 : x

x = 100/19 * 57

x = 300

5 0
3 years ago
How do you solve 36 times <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" alt="\sqrt{3}" ali
pantera1 [17]

Answer:

  62.3538

Step-by-step explanation:

There is nothing to solve. If you need a decimal value, you can use a calculator or table of square roots.

8 0
4 years ago
Help and hurry please
ira [324]

Answer:

D.

Step-by-step explanation:

First try to eliminate answers.

You have to work between 30 and 40 hours to get 1565.

It's not A or B.

Now, you can see that the choices are 31, and 35

1565 is only 15 more than 1550.

It's 31.

5 0
4 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
Which is a true statement about any two congruent chords in a circle? A.They are parallel. B.They are perpendicular. C.They form
Tom [10]
Its D. they are equidistant from the center of the circle , just took the test and that was the answer on apex
7 0
3 years ago
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