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Darina [25.2K]
3 years ago
10

Please help me with this problem I don't understand how to solve it 1.) Simplify. √18

Mathematics
2 answers:
sp2606 [1]3 years ago
7 0
4.24 :)
This is simplified and not the full answer btw
hope this helps
Citrus2011 [14]3 years ago
6 0
Answer:
3√2 or 4.2426

Explanation:
18 can be written as 9*2
This means that:
√18 can be written as √9 * √2
Now, √9 is equal to 3
So, the simplified form of √18 is 3√2 which is equal to 4.2426 in decimal form.

Hope this helps :)
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1+1+1+1+1+1+1+1+1+1+18+18+18+18+18+18+80+80+80+80+100+100+100+100+1000
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1836

Step-by-step explanation:

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Suppose your statistics professor reports test grades as​ z-scores, and you got a score of 2.492.49 on an exam. ​a) Write a sent
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Answer:

a) Option C) The score was 2.49 standard deviations higher than the mean score in the class.

b) 2.3%

Step-by-step explanation:

a) We are given that the distribution of test grades is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

\displaystyle\frac{x-\mu}{\sigma} = 2.49\\\\x - \mu = 2.49\sigma\\x = \mu + 2.49\sigma

Option C) The score was 2.49 standard deviations higher than the mean score in the class.

b) The z-score for a particular score is -2.

We have to evaluate

P(z < -2)

Calculating the value from normal z-table.

P(z < -2) = 0.023

Thus, 2.3% of  of the class scored lower than my​ friend.

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3 years ago
The Alvin Secretarial Service procures temporary office personnel for major corporations. They have found that 60% of their invo
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Answer:

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days

Step-by-step explanation:

For each invoice, there are only two possible outcomes. Either they are paid within 10 working days, or they are not. The probability of an invoice being paid within 10 working days is independent of other invoices. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

60% of their invoices are paid within ten working days.

This means that p = 0.6

A random sample of 18 invoices is checked.

This means that n = 18

What is the probability that at least 6 of the invoices will be paid within ten working days?

Either less than 6 are paid within 10 working days, or at least 6 are paid. The sum of the probabilities of these events is 1. So

P(X < 6) + P(X \geq 6) = 1

We want P(X \geq 6). So

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Then

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{18,0}.(0.6)^{0}.(0.4)^{18} \approx 0

P(X = 1) = C_{18,1}.(0.6)^{1}.(0.4)^{17} \approx 0

P(X = 2) = C_{18,2}.(0.6)^{2}.(0.4)^{16} \approx 0

P(X = 3) = C_{18,3}.(0.6)^{3}.(0.4)^{15} = 0.0002

P(X = 4) = C_{18,4}.(0.6)^{4}.(0.4)^{14} = 0.0011

P(X = 5) = C_{18,5}.(0.6)^{5}.(0.4)^{13} = 0.0045

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0 + 0 + 0 + 0.0002 + 0.0011 + 0.0045 = 0.0058

Finally

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.0058 = 0.9942

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days

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