Factor the polynomial below. 20x squared minus 45

The diameter of a circle is 9 ft. Find its area to the nearest tenth.

Drupady [299]

Step-by-step explanation:

D=9

A=pie*radius^2

=3.14*9/2*9/2

=63.58

=64

6 0

3 years ago

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Factor n^3 - n^2 + 3n - 3.

lozanna [386]

N^2(n - 1) + 3(n - 1)

(n^2 + 3)(n - 1)

the answer is c

5 0

2 years ago

20 men decided to complete a work in 30 days. but after 6 days 8 men left now in how many day the work will be completed

Serga [27]

First, we convert it into a unit like this

20 workers × 30 days = 600 units work

After leaving the 5 workers, we have 35 days to complete the work.

In 35 days only 15 workers will do the work.

So the work completed in 35 days by 15 workers = 35 ×15 =525 units

Remaining unit = 600 -525 =75 units

This unit would do by 5 workers in (75/5) = 15 days.

So after 15 days, 5 workers should leave the job.

6 0

3 years ago

What is 22% of 64.82

Phantasy [73]

Answer:

14.2604

Step-by-step explanation:

this is because 22% as a decimal would be 0.22, and since this is out of 64.82, you have to multiply 0.22 by 68.82

\\

Hope this helps <33

4 0

3 years ago

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A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for

tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of t are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.

Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

$\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29$

$s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64$

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

$\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11$

$s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08$

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a t-test for two means.

Compute the test statistic value as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

$=\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}$

$=-4.123$

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

$m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}$

$=\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}$

$=26.55\\\approx 27$

Compute the critical value for α = 0.05 as follows:

$t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052$

*Use a t-table for the values.

Thus, the critical values of t are ± 2.052.

3 0

2 years ago