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Factor out the GCF from the terms of the polynomial −2x5 − 6x3 − 16x2.

Ahat [919]

Answer:

2x^2(x^3+3x+8)

Step-by-step explanation:

Factor out the gcf which in this case is -2x^2 because that's the greatest factor all of the terms have in common this results in -2x^2(x^3+3x+8)

Btw the ^ represents the exponent

4 0

3 years ago

The data set of the diameters of the metal cylinders manufactured an automatic machine has sample size of n=29, mean of x= 49.98

Lena [83]

Answer:

we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )

Step-by-step explanation:

Given data

n=29

mean of x = 49.98 mm

S = 0.14 mm

μ = 50.00 mm

Cl = 95%

to find out

Can we be 95% confident that machine calibrated properly

solution

we know from t table

t at 95% and n -1 = 29-1 = 28 is 2.048

so now

Now for 95% CI for mean is

(x - 2.048 × S/√n , x + 2.048 × S/√n )

(49.98 - 2.048 × 0.14/√29 , 49.98 + 2.048 × 0.14/√29 )

( 49.926757 , 50.033243 )

hence we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )

8 0

3 years ago

I need help with number 8

Mumz [18]

Answer:

If the underlined digit is 5 then it is Fifty thousand (50,000)

8 0

3 years ago

Read 2 more answers

Rent and other associated housing costs, such as utilities, are an important part of the estimated costs of attendance at colleg

zhenek [66]

Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

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

where, $\bar X$ = average rent paid by 314 BYU students = $346

s = sample standard deviation = $86

n = sample of students = 314

$\mu$ = population mean monthly rent of all unmarried BYU students

So, 96% confidence interval for the population mean monthly rent, $\mu$ is ;

P(-2.082 < $t_3_1_3$ < 2.082) = 0.96

P(-2.082 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.082) = 0.96

P( $-2.082 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X -2.082 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.082 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $346 -2.082 \times {\frac{86}{\sqrt{314} } }$ , $346 +2.082 \times {\frac{86}{\sqrt{314} } }$ ]

= [335.89 , 356.10]

Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

5 0

4 years ago

Only number 8 plsssss bet no one can get this

Nesterboy [21]

Answer:

52 sq un

Step-by-step explanation:

8 0

3 years ago