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3 years ago

What is the factorization of 3x^2+27x+60

1 answer:

3 0

Factor the following:
3 x^2 + 27 x + 60
Factor 3 out of 3 x^2 + 27 x + 60:
3 (x^2 + 9 x + 20)
The factors of 20 that sum to 9 are 5 and 4. So, x^2 + 9 x + 20 = (x + 5) (x + 4):
Answer: 3 (x + 5) (x + 4)

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On the number line what is fraction is between 1/2 and 5/6

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Answer:

2/3

Step-by-step explanation:

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3 years ago

What’s a larger number is two less than four times a smaller. if the sum of the number is 18

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Let x be the larger number and y be the smaller number.
Using the given info, we get
x = 4y - 2
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Enter this value into x = 4y - 2
x = 4(4) - 2
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3 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

. You are interested in the amount of time teenagers spend weekly working at part-time jobs. A random sample of 15 teenagers was

agasfer [191]

Answer:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

$\bar X=147.3$ represent the sample mean for the amount of time spent at part time jobs

$s=50$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =120$ represent the value to check

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

4 0

3 years ago