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2 years ago

What is the factored form of 6n4 – 24n3 + 18n?

Mathematics

2 answers:

seropon [69]2 years ago

8 0

Answer: Its factored form will be

$6n(n-1)(n^2-3n-3)$

Step-by-step explanation:

Since we have given that

$6n^4-24n^3+18n$

So, we need to factor ,

$6n(n^3-4n^2+3)$

Now,

$\text{Let }p(n)=n^3-4n^2+3$

Now, by hit and trial method, we put n=1,

$p(1)=1^3-4\times1^2+3=1-4+3=1-1=0$

$n=1\\n-1=0$

So, n-1 is also factor of p(n).

Now,

$\frac{n^3-4n^2+3}{n-1}=n^2-3n-3$

So, its factored form will be

$6n(n-1)(n^2-3n-3)$

Oduvanchick [21]2 years ago

3 0

Answer:

6n(n^3-4n^3+3) or C on e2020

Step-by-step explanation:

The GCF is 6n. divide all the terms by 6n.

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The library is 1.75 miles directly north from the school. The park is 0.6 miles directly south of the school. How far away is th

Kryger [21]

Answer:

2.35 miles

Step-by-step explanation:

Please find the attachment for visual understanding.

We are told that the library is 1.75 miles directly north from the school and the park is 0.6 miles directly south of the school.

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$(1.75+0.6)\text{ miles}=2.35\text{ miles}$

Therefore, library is 2.35 miles away from park.

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2 years ago

At a construction site, a builder plans to cut down an 8-foot-by-9-foot piece of plywood. She plans to discard any pieces that s

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Step-by-step explanation:

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2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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2 years ago

Convert to the nearest tenth. gal. = 319 qt.

lisov135 [29]

The answer is 79.75 converted to 80.0
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1 year ago

A wedding planner purchased small and large lanterns for a wedding reception. The small lanterns cost $25 each, and the large la

Marina CMI [18]

Answer: y=12 x=28

Explanation:
x=Amount of Small Lanterns y=Amount of Large Lanterns
25x+40y=1180
x+y=40
x=40-y
25(40-y)+40y=1180
1000-25y+40y=1180
1000+15y=1180
15y=180
y=12
x+12=40
x=28

5 0

2 years ago

Read 2 more answers