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3 years ago

Simplify each expression by distributing and combining like terms 6(x + 1) – 5(x + 2)

Mathematics

2 answers:

Vlada [557]3 years ago

5 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{x - 9}}}}}$

Step-by-step explanation:

$\star \: \sf{6( x + 1) - 5(x + 2)}$

Distribute 6 through the parentheses

Similarly, distribute -5 through the parentheses

$\dashrightarrow{ \sf{6x + 6 - 5x - 10}}$

Collect like terms

Like terms are those which have the same base

$\dashrightarrow{ \sf{6x - 5x + 6 - 10}}$

$\dashrightarrow{ \sf{x + 6 - 10}}$

The negative and positive integers are always subtracted but posses the sign of the bigger integer

$\dashrightarrow{ \boxed{ \sf{x - 4}}}$

Hope I helped!

Best regards! :D

Molodets [167]3 years ago

3 0

Answer:

x-4

Step-by-step explanation:

6(x+1)-5(x+2)

=6x+6-5x-10

=x-4

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The mean shoe size of the students in a math class is 7. 5. Most of the shoe sizes fall within 1 standard deviation, or between

lana [24]

Answer: 1.5

Step-by-step explanation:Let the standard deviation of the shoe size be represented by ' ' and the mean shoe size be represented by ' = 7.5'. Also, the sizes of the shoe be represented by ' '. So, . Hence, the standard deviation of the shoe size data for the math class is 1.5.

4 0

2 years ago

Write an equation of the line containing the given point and parallel to the given line.

Dafna11 [192]

Answer:

y = 5/9x - 37/3

Step-by-step explanation:

5x - 9y = 8

-9y = -5x + 8

y = 5/9x - 8/9

When a line is parallel, the slope remains the same.

y = 5/9x + b

-9 = 5/9(6) + b

-9 = 10/3 + b

-37/3 = b

6 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 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:

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