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3 years ago

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Mathematics

1 answer:

MAVERICK [17]3 years ago

6 0

Answer: 9a^4 - b^2

= (3a^2)^2 - b^2

= (3a^2 + b)(3a^2 - b) using a^2 - b^2 = (a+b)(a-b)

hope it helps.

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Answer:

Step-by-step explanation:

The fast way: Testing the options!

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Read 2 more answers

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.

From the central limit theorem we have the formula:

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

a) $z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40$

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

$\therefore P(\bar X\:>\:1224)=0.0082$

b) $z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00$

The area to the left of z=3.00 is 0.9987

$\therefore P(\bar X\:$

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

$z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40$

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

$P(1200\:$

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) $z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65$

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

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