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ss7ja [257]
3 years ago
14

In a wildlife park, there were 18 gray wolves. After three years, there were 27 gray wolves. What is the percent gain?

Mathematics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

50 percent gain

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I need the percentage of the shaded area.
arsen [322]

Answer:

1.22

Step-by-step explanation:

Since there is one whole shaded. And 2 tens, and 2 ones. Its 1.22.

3 0
3 years ago
You have already run 4 miles. If you run at a speed of 8 miles per hour, how many total miles will you run in 2 more hours? Choo
Inessa [10]

Answer:

20 miles

Step-by-step explanation:

I'm not sure if that is exactly how you solve it but

If its

8x+4 as the equation and x is the number of hours run

the total number of miles run should be 20 miles

8(2)+4=20

8 0
2 years ago
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3x3-5/5x6+19 i don’t know what I’m don’t wrong
KonstantinChe [14]

Hi there!

For these two equations, we would be using PEMDAS...

3·3=9-5=4

5·6=30+19=49

Hope this helps!

4 0
3 years ago
A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w
melamori03 [73]

Answer:

The hypothesis is:

<em>H₀</em>: p_{X}-p_{Y}=0.

<em>Hₐ</em>: p_{X}-p_{Y}.

Step-by-step explanation:

Let <em>X</em> = number of men who exercise regularly and <em>Y</em> = number of women who exercise regularly.

The information provided is:

n_{X}=150\\X=88\\n_{Y}=200\\Y=130

Compute the sample proportion of men and women who exercise regularly as follows:

\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587

\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and \hat p_{X}=0.587.

The random variable <em>Y</em> also follows a Binomial distribution with parameters <em>n</em> = 200 and \hat p_{Y}=0.65.

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

\hat p=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion <em>z</em>-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

<em>H₀</em>: The proportion of women is same as men who exercise regularly, i.e. p_{X}-p_{Y}=0.

<em>Hₐ</em>: The proportion of women is more than men who exercise regularly, i.e. p_{X}-p_{Y}.

6 0
3 years ago
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In a video game, the chance of rain each day is always 30%. At the beginning of each day in the video game, the computer generat
Scorpion4ik [409]

Answer:

I would divide 50 by 4 (Because there are usually 4 weather conditions in game) and whichever the number is closest to it would pick that weather condition

8 0
2 years ago
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