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iogann1982 [59]
3 years ago
7

Can anyone please help me out? I have to find the Greatest Common Factor for each, thank you!​

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

1= 4x^2

2= 5b^4

3= 3u^2v^2w

4= 7

5= bc^3d^6

Note: ^ stands for exponents

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Yep here we go again:
Alchen [17]

Answer:

126

Step-by-step explanation:

360 x 35%= 126

360 x 0.35 = 126

3 0
2 years ago
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HELP PLEASE 25 POINTS
erastova [34]

Answer:

$15, $8, $1, respectively

Step-by-step explanation:

Line up the number of years with the line and check the pay.

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3 years ago
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An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0
3 years ago
IHELP if 1 out of 5 animals are cats then how many cats are there in an ecosystem of 200 animals???​
mote1985 [20]
There are 40 cats in the ecosystem.

You set 1/5 equal to x/200. Then you cross multiply the fractions to get 5x=200. You divide both terms by 5 to get x, which equals 40.

3 0
3 years ago
7. A clothing trunk is 30 inches tall, 48 inches wide,
Vinvika [58]

Answer:

34,560 in^3

Step-by-step explanation:

A clothing trunk is 30 inches tall, 48 inches wide, 24 inches deep

The amount is cubic feet the trunk will hold can be calculated as follows

= 30×48×25

= 34,560 in^3

4 0
2 years ago
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