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3 years ago

Which expression is equivalent to 9q^2 - 2/3 (3q - 7) + 5q^2?

Mathematics

1 answer:

nikitadnepr [17]3 years ago

3 0

Answer:

14q^2−2q+14/3

Step-by-step explanation:

9q^2 - 2/3 (3q - 7) + 5q^2

Use the distributive property to multiply -2/3 by 3q−7.

9q^2 - 2q + 14/3 + 5q^2

Combine 9q^2 and 5q^2 to get 14q^2

14q^2−2q+14/3

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Factor the expression completely <img src="https://tex.z-dn.net/?f=%20%7B%20%5Ctimes%20%7D%5E%7B2%7D%20%20-%202x%20-%2063%20"

Ainat [17]

Use form x^2 + bx + c
b = -2
c = -63
find a pair of integers whose product equals c and sum equals b

so, find a product that equals -63 and a sum that equals -2

-9 + 7 = -2
-9 • 7 = -63

write in factored form ->
(x - 9)(x + 7)

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3 years ago

HELP ASAP!

VashaNatasha [74]

Answer: D) -3x2-5x-15

Step-by-step explanation:

-4x2-5(x+3)+x2

-3x2-5x-15

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3 years ago

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Step-by-step explanation:

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3 years ago

Connie has saved up $15 to purchase a new CD from the local store. The sales tax in her county is 5% of the sticker price. Write

bonufazy [111]

X + 0.05x = 15
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3 years ago

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Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random

Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = $50,000

$\sigma$ = population standard deviation = $25,000

n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P( $\bar X$ > $52,000)

P( $\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

= 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0

3 years ago