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Diano4ka-milaya [45]
3 years ago
7

4^3 · 4-^6 Simplified A) 4-3 B) 4-9 C) 4-18 D) 43

Mathematics
2 answers:
Triss [41]3 years ago
8 0

4^3 · 4-^6

= 4^(3 + (-6))

= 4^-3

Answer

A) 4^-3

zavuch27 [327]3 years ago
3 0

the answer is a......



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(5 Points)<br> all (1 – 3K)<br> -396<br> 0-374<br> 0 -198<br> 0 -187
Triss [41]

Answer:−3K−395,−374,−198,−187

Step-by-step explanation:

Remove parentheses.

1−3K−396,0−374,0−198,0−187

Simplify  1-3K-3961−3K−396  to  -3K-395−3K−395.

−3K−395,0−374,0−198,0−187

Simplify  0-3740−374  to  -374−374.

−3K−395,−374,0−198,0−187

Simplify  0-1980−198  to  -198−198.

−3K−395,−374,−198,0−187

Simplify  0-1870−187  to  -187−187.

−3K−395,−374,−198,−187

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3 years ago
Zoe supplies costumes to a number of theater companies. She
KengaRu [80]

Answer:3

Step-by-step explanation:

6 0
2 years ago
Can sb please answer this
Fudgin [204]

Answer:

A

Step-by-step explanation:

A line of best fit needs to match the direction of the points and fit in-between the points as much as possible

C is not the right direction

D is not the right direction

B is to the left of the points

3 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 <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
3 years ago
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Flura [38]
Equation: 18(20) * 7 or 18(20) x 7
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