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kirill115 [55]
3 years ago
11

Solve : 4+3x =(3x-7)+9.

Mathematics
2 answers:
lesantik [10]3 years ago
6 0
X=3
Hope this helped
dezoksy [38]3 years ago
5 0

Answer:

D.) x = 3

Step-by-step explanation:

4 + 3x = 2(3x - 7) + 9

4 + 3x = 6x - 14 + 9

3x - 6x = 9 - 14 - 4

-3x = -9

x = -(-9)/3

x = 9/3

x = 3

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B, C, and E

Step-by-step explanation:

<u>Simplify all answer choices:</u>

A.

5(2x+10)+1=

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10x+51

B.

7(x+2)+3x-3=

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10x+11

C.

3(3x+4)+x-1=

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D.

2(6x+4)+2x+5=

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Answer choices B, C, and E simplify to 10x+11

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3 years ago
The average weight of the top 5 fish at a fishing tournament was 13.6 pounds. Some of the weights of the
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3 years ago
Write an integer whose absolute value is greater than itself.
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<span>The set of an integer is called the set Z
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8 0
3 years ago
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

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

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

(b) The test statistic is 3.571.

(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}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

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;

                               T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

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

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

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

8 0
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