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Anna [14]
3 years ago
11

What are the zeros of the polynomial function f(x) = x3 - 2x2 - 24x?

Mathematics
2 answers:
emmainna [20.7K]3 years ago
4 0

Answer:


Step-by-step explanation:

first factor out a variable: x(x^2-2x-24)

now solve the quadratic: x(x-6)(x+4)

When you set these expressions equal to zero, you get 0, 6, and -4 as your answers.


Nataliya [291]3 years ago
3 0
First factor out a variable: x(x^2-2x-24)
now solve the quadratic: x(x-6)(x+4)
When you set these expressions equal to zero, you get 0, 6, and -4 as your answers.
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Simplify:<br> k-10k+1 +<br> 1<br> 5k-2k-3<br> k-1
Ne4ueva [31]

Answer:

30k + 1 -1

Step-by-step explanation:

3 0
3 years ago
1. Write a number whose square is between 200 and 300.
Nadya [2.5K]
15

Square root of 200 = 14.14
Square root of 300 = 17.32

15^2 = 225
8 0
3 years ago
Joel has 24 sports trophies. Of the trophies 1/8 are soccer trophies. How many soccer does Joel have ?
tigry1 [53]

Answer:

We are given that Joel has 24 sports trophies, of the 24 trophies there are 1/8 soccer trophies.


1/8 of 24 is 3.

24 divided by 8 = 3 soccer trophies

There are 3 soccer trophies all the rest of the trophies, which is 21 trophies, will be other sports. 21 trophies + 3 soccer trophies = 24 trophies in total



Step-by-step explanation:

Have a great rest of your day
#TheWizzer

3 0
2 years ago
A small hair salon in Denver, Colorado, averages about 82 customers on weekdays with a standard deviation of 16. It is safe to a
Natali5045456 [20]

Answer:

There is a 3.44% probability to get a sample average of 93 or more customers if the manager had not offered the discount.

Step-by-step explanation:

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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A small hair salon in Denver, Colorado, averages about 82 customers on weekdays with a standard deviation of 16. This means that \mu = 82, \sigma = 16

She reports that her strategy has worked since the sample mean of customers during this 7 weekday period jumps to 93. What is the probability to get a sample average of 93 or more customers if the manager had not offered the discount?

This is 1 subtracted by the pvalue of Z when X = 93.

We also have to find a standard deviation of the sample(that is, the 7 days), so:

s = \frac{\sigma}{\sqrt(7)} = \frac{16}{\sqrt(7)} = 6.05

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

Z = \frac{93-82}{6.05}

Z = 1.82

Z = 1.82 has a pvalue of 0.9656.

This means that there is a 1-0.9656 = 0.0344 = 3.44% probability to get a sample average of 93 or more customers if the manager had not offered the discount.

6 0
3 years ago
Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance. For what fraction of
Kruka [31]

Answer:

Lola was in lead for \frac{6}{25} of a mile.

Step-by-step explanation:

We have been given that Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance.

To find the fraction of a mile for which Lola was in the lead, we need to find 4/5 of 3/10 as:

\text{Fraction for which Lola was in lead}=\frac{3}{10}\times\frac{4}{5}

\text{Fraction for which Lola was in lead}=\frac{3}{5}\times\frac{2}{5}

\text{Fraction for which Lola was in lead}=\frac{3\times2}{5\times5}

\text{Fraction for which Lola was in lead}=\frac{6}{25}

Therefore, Lola was in lead for \frac{6}{25} of a mile.

7 0
3 years ago
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