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2 years ago

Factor completely 3x^2+9x+6

Mathematics

1 answer:

anastassius [24]2 years ago

7 0

Answer:

Step-by-step explanation:

3x²+9x+6

=3(x²+3x+2)

=3[x²+x+2x+2]

=3[x(x+1)+2(x+1)]

=3(x+1)(x+2)

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Which expression is equivalent to negative 32 5/3?

Sloan [31]

Im confused. Its not equal to any of the choices

Unless C is the cube root of 32^5.

[t3] \sqrt[n]{x} [32^5]

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3 years ago

what in a equation to represent the relationship "a number decreased by negative 7 and is the same thing as 48

murzikaleks [220]

I think it would be n-(-7)=48

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3 years ago

Patty is ordering vests for her scout troop to wear in the Veterans Day parade. She ordered 12 vests from the Festive Features c

anygoal [31]

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2 years ago

Which is greater .836 or .83

anygoal [31]

Just do like this
.836
.83
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3 years ago

Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th

babymother [125]

Answer:

$\= x = 40.85$

$E = 5.85$

$t_c = 2.08$

$t_c = 1.282$

Explanation:

From the question we are told that

The sample size is n = 100

The upper limit of the 95% confidence interval is b = 47.2 years

The lower limit of the 95% confidence interval is a = 34.5 years

Generally the sample mean is mathematically represented as

$\= x = \frac{a + b }{2}$

=> $\= x = \frac{47.2 + 34.5 }{2}$

=> $\= x = 40.85$

Generally the margin of error is mathematically represented as

$E = \frac{b- a }{ 2}$

=> $E = \frac{47.2- 34.5 }{ 2}$

=> $E = 5.85$

Considering question C a

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

The sample size is n = 22

Given that the sample size is not sufficient enough i.e $n < 30$ we will make use of the student t distribution table

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 22 - 1$

=> $df = 21$

Generally from the student t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of 21 is

$t_c =t_{\frac{\alpha }{2} , 21 } = 2.08$

Considering question C b

From the question we are told the confidence level is 80% , hence the level of significance is

$\alpha = (100 - 80 ) \%$

=> $\alpha = 0.20$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$t_c =Z_{\frac{\alpha }{2} } = 1.282$

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2 years ago