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3 years ago

Apply the distributive property to factor out the greatest common factor. 45e − 27 f =

Mathematics

2 answers:

Varvara68 [4.7K]3 years ago

6 0

$45e=9\cdot5e\\\\27f=9\cdot3f\\\\45e-27f=9\cdot5e-9\cdot3f=9\cdot(5e-3f)$

jeyben [28]3 years ago

3 0

Answer:

45e − 27 f = 9(5e - 3f)

Step-by-step explanation:

We are given a expression:

45e-27f

We have to apply the distributive property to factor out the greatest common factor.

Distributive property says that:

a(b+c)=ab+bc

45e-27f

=9.5.e-9.3.f

Taking 9 common from both the terms

=9(5e-3f)

Hence, 45e − 27 f = 9(5e - 3f)

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4 years ago

A population of values has a normal distribution with μ = 247 and σ = 62.2. You intend to draw a random sample of size n = 16. (

nadya68 [22]

Answer:

a) $z = \frac{295.2-247}{62.2}=0.772$

And using the normal distribution table or excel we got:

$P(Z>0.772) = 1-P(Z$

b) $z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119$

And we can use the normal standard table or excel in order to find the probability and we got:

$P(z>3.119) =1-P(Z$

Step-by-step explanation:

For this case we know that the random variable of interest is normally distributed with the following parameters:

$X \sim N (\mu = 247, \sigma =62.2)$

Part a

We want to find this probability:

$P(X>295.2)$

And we can use the z score formula given by:

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

Replacing we got:

$z = \frac{295.2-247}{62.2}=0.772$

And using the normal distribution table or excel we got:

$P(Z>0.772) = 1-P(Z$

Part b

We select a random sample of size n = 16 and we try to find this probability:

$P(\bar X >295.2)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119$

And we can use the normal standard table or excel in order to find the probability and we got:

$P(z>3.119) =1-P(Z$

8 0

3 years ago