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m_a_m_a [10]
3 years ago
12

Can Someone Help With This Question Please

Mathematics
2 answers:
8090 [49]3 years ago
8 0

4x²+x³

Factor is out x³ expression

The answer is :

{x}^{2}  \times (4 + 3x)

Carry learning!

Dovator [93]3 years ago
7 0
<h3><u>Given </u><u>:</u><u>-</u></h3>

  • We have given expression, 4x² + x³

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • We have to factorise this expressions as fully as possible

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

4x² + x³

  • Arrange it accordingly

x³ + 4x²

  • X is common in the expression

x²( x + 4 )

  • x² ( 4 + x)

Hence, After factorization we got two possible factors of the above expression that is x² and (4 + x )

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andrey2020 [161]
9 + x > (equal to or greater than) 15
6 more students need to sign up because 15-9=6
3 0
3 years ago
(-33)+(-7)=
sasho [114]

Answer:

-40

-6

51

Step-by-step explanation:

-33+-7 you are adding two negatives

9+(-15) you are adding a negative to a positive which gets you closer to zero

5-(-46) you are subtracting a negative which is adding

7 0
3 years ago
Use the Central Limit Theorem to find a mean given a probability Question A video game company sells an average of 132 games a m
aev [14]

Answer:

The number of games must a store sell in order to be eligible for a reward is 135.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of video games sold in a month by the sores.

The random variable <em>X</em> has a mean of, <em>μ</em> = 132 and a standard deviation of, <em>σ</em> = 9.

It is provided that the company is looking to reward stores that are selling in the top 7%.

That is, P (\bar X > \bar x) = 0.07.

The <em>z</em>-score related to this probability is, <em>z</em> = 1.48.

Compute the number of games must a store sell in order to be eligible for a reward as follows:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}

\bar x=\mu+z\cdot \sigma/\sqrt{n}

   =132+1.48\times (9/\sqrt{36})\\\\=132+2.22\\\\=134.22\\\\\approx 135

Thus, the number of games must a store sell in order to be eligible for a reward is 135.

5 0
3 years ago
A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma
IrinaVladis [17]

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when M = 0.08

We have that \sigma = 0.72

M = z*\frac{\sigma}{\sqrt{n}}

0.08 = 2.575*\frac{0.72}{\sqrt{n}}

0.08\sqrt{n} = 2.575*0.72

\sqrt{n} = \frac{2.575*0.72}{0.08}

(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}

n = 537.1

Rounding up

538 books should be tested.

6 0
3 years ago
Bobby had 36 books in his locker. Some were library books, some were textbooks, and the rest were telephone books. The number of
emmainna [20.7K]
Make variables for each
L = library books
P = telephone books
T = text books

so.... L+P = 2T 
now we have to fill in numbers
8 0
4 years ago
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