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3 years ago

Factor out a greatest common factor 25x^3+10x^2+5x

Mathematics

2 answers:

slava [35]3 years ago

8 0

Answer:

Step-by-step explanation:

Dmitriy789 [7]3 years ago

7 0

The greatest common factor would be 5x because all the variables can be divided by 5x.

The answer itself already factored would be

5x(5x^2+2x+1)

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Maria cut 3/4 yard of ribbon into pieces that are 1/8 yard long. How many pieces of ribbon did Maria cut? don't be a b.*.t.c.h.

Akimi4 [234]

<h3>Answer: 6 pieces</h3>

===============================================

Explanation:

3/4 = 6/8 after multiplying top and bottom of the fraction by 2

She has 6/8 of a yard of ribbon.

If she cuts the ribbon into pieces of 1/8 of a yard each, then she has 6 pieces overall

We can see this if we solve the equation (1/8)x = 6/8. Multiply both sides by 8 to get x = 6.

8 0

3 years ago

I need help, I did 1-2b, but i do not mind someone answering it either way so I can double check, but I am mainly stuck with 2c

garri49 [273]

Answer:

i belive the answer is c because they always try and trick you making you choose the wrong one and all the evidence cites that it could be c!!

Step-by-step explanation:

3 0

2 years ago

M∠ABD=87°<br> m∠ABC=9x−1°<br> m∠CBD=6x+58 <br><br> Encuentra m∠ABC

Softa [21]

Answer: 9x - 1 = 9(2) - 1 = 17

Step-by-step explanation:

m∠ABD = m∠ABC + m∠CBD

87 = 9x -1 + 6x + 58

87 = 15x + 57

30 = 15x

2 = x

5 0

2 years ago

Which terms must be removed from 1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 if the sum of the remaining terms is to equal 1?

liraira [26]

1/4

Step-by-step explanation:

0.5+0.25+0.17+0.13+0.1+0.1=1.25

4 0

2 years ago

For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it

amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average mpg = 15.6 mpg

s = sample standard deviation = 1.9 mpg

n = sample of minivans = 100

$\mu$ = population average mpg

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.987 < $t_9_9$ < 1.987) = 0.95 {As the critical value of t at 99 degree

of freedom are -1.987 & 1.987 with P = 2.5%}

P(-1.987 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.6-1.987 \times {\frac{1.9}{\sqrt{100} } }$ , $15.6+1.987 \times {\frac{1.9}{\sqrt{100} } }$ ]

= [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0

2 years ago