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3 years ago

9

a small Computing Center has found that the number of jobs submitted per day to its computers has a mean of 65 jobs and a standa

rd deviation of 7.where do we expect the number of jobs per day to fall most of the time(95.4%)?

1 answer:

jek_recluse [69]3 years ago

6 0

Check the attached file for the solution.

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aalyn [17]

3.141592653589793238462643383279 502884197169399375105820974944 5923078164062862089986280348253421170679

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3 years ago

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The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. Wh

Alla [95]

Answer:

2.3%

Step-by-step explanation:

The formula to convert x into z distribution is

$z=\frac{x-\mu}{\sigma}$

Where z is the test statistic

x is what we are looking for (in this case 37)

$\mu$ is the mean (in this case, it is 25)

$\sigma$ is the standard deviation (we have 6)

<em>Plugging these into the formula, we get:</em>

$z=\frac{x-\mu}{\sigma}\\z=\frac{37-25}{6}\\z=2$

Thus we can say we want $P(z\geq 2)$

<u>Note: </u> $P(z\geq a)=1-P(z\leq a)$

The table given is for any z where $P(z\leq a)$

Thus, now we have:

$P(z\geq 2)\\=1-P(z\leq 2)\\=1-0.9772\\=0.0228\\$

0.0228 into percentage is 0.0228 * 100 = 2.28%

Rounded, we get 2.3%

Third answer choice is right.

3 0

4 years ago

Help!!<br><br> which expression is equivalent to (m^5 n /pq^2) ^4

lana66690 [7]

$\text{Use:}\\\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{nm}\\\\......................................................\\\\\left(\dfrac{m^5n}{pq^2}\right)^4=\dfrac{(m^5n)^4}{(pq^2)^4}=\dfrac{(m^5)^4n^4}{p^4(q^2)^4}=\dfrac{m^{20}n^4}{p^4q^8}$

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3 years ago

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

a.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

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3 years ago

What is the factored form of the expression 9x^2 + 6x +1

liraira [26]

Answer:

(3x+1)^2

Step-by-step explanation:

4 0

3 years ago

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