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2 years ago

12

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

s in pounds.

1 answer:

77julia77 [94]2 years ago

4 0

Answer:

x = 2 1/4 or 2.25

Step-by-step explanation:

to start we can make an equation

9lbs = 4x

divide by 4

9/4 = x

x = 2.25 or 2 1/4

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2. Randy has a house that is worth $250,000 but he still owes $175,000. He has a car that is worth $50,000 but he still owes $37

faust18 [17]

Answer:

$91,200

Step-by-step explanation:

I gave it my best shot sorry if it wrong

3 0

2 years ago

Please help! I mark brainliest :3

vladimir1956 [14]

no for a
yes for B
no for c
that what I think It is opinion could be wrong

3 0

3 years ago

Use prime factorization to reduce the fraction 36/180

laila [671]

Answer:

Step-by-step explanation:

36 = 6* 6 = 2* 3 * 2* 3 = 2*2*3*3

180 = 2 * 2 * 3 * 3 * 5 = 2 * 2 * 3*3*5

$\frac{36}{180}= \frac{2*2*3*3}{2*2*3*3*5} = \frac{1}{5}$

Factors that are common will be cancelled

5 0

2 years ago

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 <em>t</em> 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:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 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 <em>t</em>-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 <em>α</em> = 0.05 as follows:

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

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0

2 years ago

Show that the triangles are similar by comparing the ratios of the corresponding sides. Simplify your answer completely in order

kirill [66]

Answer/Step-by-step explanation:

AC = 1.2

AB = 4

BC = 2.6

DF = 3

DE = 10

EF = 6.5

Thus:

$\frac{DE}{AB} = \frac{10}{4} = \frac{5}{2}$

$\frac{DF}{AC} = \frac{3}{1.2} = \frac{3*10}{1.2*10} = \frac{30}{12} = \frac{5}{2}$

$\frac{EF}{BC} = \frac{6.5}{2.6} = \frac{6.5*10}{2.6*10} = \frac{65}{26} = \frac{5}{2}$

The ratio of their corresponding sides are all equal to ⁵/2. Therefore, both triangles are similar.

4 0

3 years ago