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s344n2d4d5 [400]
3 years ago
12

How do i factor x(3x+5) - 4(3x+5)

Mathematics
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

(3x + 5)(x - 4)

Step-by-step explanation:

Given

x(3x + 5) - 4(3x + 5) ← factor out (3x + 5) from each term

= (3x + 5)(x - 4) ← in factored form

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What is the slope of (4,1) and (6,-1)
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Answer:

slope = -1

Step-by-step explanation:

you <u>subtract the y-values</u> (-1-1=-2) then <u>subtract the x-values</u> (6-4=2) then divide the sum of the y-values by the sum of the x-values

\frac{-2}{2} which simplifies to a slope of -1

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Figure ABCD is transformed to figure A prime B prime C prime D prime, as shown: A coordinate grid is shown from negative 5 to 0
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The lowest common denominator for the fractions 8/64 and 8/32 is what
KiRa [710]

the least common denominator for 32 and 64 is

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3 years ago
PLS HELP ASAP ILL GIVE BRAINLKEST PLS THANKS
Licemer1 [7]

Answer:

6 and 28

Step-by-step explanation:

Question 8: Since the sum is atmost 16, the largest possible value of the number is available when the sum is 16. Then, we have 2x + 4 = 16, and that equals 2x = 12, which leads to x = 6.

Question 9. Since your bank value has to be above 525, and 814 dollars keep decreasing 10 each day, we know that it will stop at 534, because having 524 dollars will be below 525. Now, 814 - 534 = 280, and since you pay 10 every day, 280/18 = 28 days.

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Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic t.
lawyer [7]

Using the t-distribution, it is found that the p-value of the test is 0.007.

At the null hypothesis, it is <u>tested if the mean lifetime is not greater than 220,000 miles</u>, that is:

H_0: \mu \leq 220000

At the alternative hypothesis, it is <u>tested if the mean lifetime is greater than 220,000 miles</u>, that is:

H_1: \mu > 220000.

We have the <u>standard deviation for the sample</u>, thus, the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem:

\overline{x} = 226450, \mu = 220000, s = 11500, n = 23

Then, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{226450 - 220000}{\frac{11500}{\sqrt{23}}}

t = 2.69

We have a right-tailed test(test if the mean is greater than a value), with <u>t = 2.69</u> and 23 - 1 = <u>22 df.</u>

Using a t-distribution calculator, the p-value of the test is of 0.007.

A similar problem is given at brainly.com/question/13873630

8 0
2 years ago
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