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Maru [420]
3 years ago
11

14. 9(y - 4) - 7y = 5(3y - 2step by step please​

Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
8 0

Answer:

y=-2

Step-by-step explanation:

9(y−4)−7y=5(3y−2)

(9)(y)+(9)(−4)+−7y=(5)(3y)+(5)(−2)

9y+−36+−7y=15y+−10

(9y+−7y)+(−36)=15y−10

2y+−36=15y−10

2y−36=15y−10

Step 2:

2y−36−15y=15y−10−15y

−13y−36=−10

Step 3:

−13y−36+36=−10+36

−13y=26

divide

/13     /13

y=-2

ludmilkaskok [199]3 years ago
3 0

Answer:

y=-2

Step-by-step explanation:

Simplifying

9(y + -4) + -7y = 5(3y + -2)

Reorder the terms:

9(-4 + y) + -7y = 5(3y + -2)

(-4 * 9 + y * 9) + -7y = 5(3y + -2)

(-36 + 9y) + -7y = 5(3y + -2)

Combine like terms: 9y + -7y = 2y

-36 + 2y = 5(3y + -2)

Reorder the terms:

-36 + 2y = 5(-2 + 3y)

-36 + 2y = (-2 * 5 + 3y * 5)

-36 + 2y = (-10 + 15y)

Solving

-36 + 2y = -10 + 15y

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '-15y' to each side of the equation.

-36 + 2y + -15y = -10 + 15y + -15y

Combine like terms: 2y + -15y = -13y

-36 + -13y = -10 + 15y + -15y

Combine like terms: 15y + -15y = 0

-36 + -13y = -10 + 0

-36 + -13y = -10

Add '36' to each side of the equation.

-36 + 36 + -13y = -10 + 36

Combine like terms: -36 + 36 = 0

0 + -13y = -10 + 36

-13y = -10 + 36

Combine like terms: -10 + 36 = 26

-13y = 26

Divide each side by '-13'.

y = -2

Simplifying

y = -2

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Step-by-step explanation:

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Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

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(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

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The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

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             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

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Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

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