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Scrat [10]
3 years ago
15

Simplify completely quantity 2 x squared plus 20 x plus 32 all over x squared minus 2 x minus 80

Mathematics
2 answers:
VARVARA [1.3K]3 years ago
4 0

Answer:  Simplified form will be

\frac{2(x+2)}{(x-10}

Explanation:

Since we have given that

\frac{2x^2+20x+32}{x^2-2x-80}

and we need to simplify it,

First we take 2 as common factor from the numerator,

So, it becomes,

\frac{2(x^2+10x+16)}{x^2-2x-80}

By using splitting the middle terms, we get

\frac{2(x^2+8x+2x+16)}{x^2-10x+8x-80}\\\\=\frac{2[(x(x+8)+2(x+8)]}{x(x-10)+8(x-10)}\frac{2(x+8)(x+2)}{(x-10)(x+8)}\\\\=\frac{2(x+2)}{(x-10}

Hence, Simplified form will be

\frac{2(x+2)}{(x-10}



erastovalidia [21]3 years ago
3 0
The answer is 2(x+2)/x-10
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The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
4 years ago
Please help me please will give brainliest to ​
uranmaximum [27]

Answer:

B.

<h3>step by step explanation</h3>

I hope it's help

7 0
3 years ago
10) Thirty-seven percent of the American population has blood type O+. What is the probability that at least four of the next fi
lisabon 2012 [21]

Answer:

0.06597

Step-by-step explanation:

Given that thirty-seven percent of the American population has blood type O+

Five Americans are tested for blood group.

Assuming these five Americans are not related, we can say that each person is independent of the other to have O+ blood group.

Also probability of any one having this blood group = p = 0.37

So X no of Americans out of five who were having this blood group is binomial with p =0.37 and n =5

Required probability

=The probability that at least four of the next five Americans tested will have blood type O+

= P(X\geq 4)\\= P(X=4)+P(x=5)\\= 5C4 (0.37)^4 (1-0.37) + 5C5 (0.37)^5\\= 0.06597

7 0
3 years ago
Read 2 more answers
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