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olya-2409 [2.1K]
2 years ago
13

(3x-4)(x^2-5x-1) (3x−4)(x 2 −5x−1)

Mathematics
1 answer:
max2010maxim [7]2 years ago
5 0

Answer:

3x 3 - x 2 - 5x + 3

Step-by-step explanation:

To find the products, multiply x by each element of the longer expression 3x 2 + 2x - 3.

Then multiply each element of the longer expression again by -1 and add the results of the two products.

(3x 2 + 2x - 3)(x - 1)

= 3x^3 + 2x^2 - 3x - 3x^2 - 2x +3

Rearrange = 3x^3 + 2x^2 - 3x^2 - 3x - 2x +3

simplify = 3x^3 - x^2 - 5x + 3

will have a good Day!

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you and your family are making a trip to see a friend.you drive 56 miles per hour for the first 3 hours and then drive 63 miles
Aleonysh [2.5K]
56 x 3 = 168

63 x 5 = 315

315 + 168 = 483

483 miles
4 0
4 years ago
The cost for bowling includes $4.00 for shoes and $4.75 for each
Irina-Kira [14]

Answer:

y = 4.75x + 4

Step-by-step explanation:

it would be $4.75 for each game and you would only pay for shows once

8 0
3 years ago
Read 2 more answers
Please write it down step by step :)
grandymaker [24]

Answer:

-6

Step-by-step explanation:

Some nasty order of operations coming up.

Firstly, deal with that squared:

-12 / 3 * (-8 + 16 - 6) + 2

Simplify the bracket:

-12 / 3 * 2 + 2

Simplify -12 / 3:

-4 * 2 + 2

Simplify -4 * 2:

-8 + 2

Simplify:

-6

3 0
3 years ago
Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.
Fynjy0 [20]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10.

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when X = 500. So

Z = \frac{X - \mu}{s}

Z = \frac{500 - 465}{10}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

8 0
3 years ago
What is the mode of this data set?<br><br> {7, 10, 12, 9, 12, 4, 13, 14}
trapecia [35]
The mode of this set of data is 12
3 0
3 years ago
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