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masya89 [10]
3 years ago
7

Solve for v. -8(v+7)=3v-23 Simplify your answer as much as possible. v=

Mathematics
2 answers:
lozanna [386]3 years ago
6 0

Answer:

v=-3

Step-by-step explanation:

-8v-56=3v-23 (brackets expanded)

-56+23=3v+8v (like termz together)

-33=11v (simplified according to maths operation )

-3=v

aksik [14]3 years ago
3 0

Answer:

v = -3

Step-by-step explanation:

-8(v + 7) = 3v - 23

Expand.

-8v - 56 = 3v - 23

Add -3v and 56 on both sides.

-8v - 56 + 56 - 3v = 3v - 23 + 56 - 3v

-8v - 3v = -23 + 56

Combine like terms.

-11v = 33

Divide -11 into both sides.

-11v/-11 = 33/-11

v = -3

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helpp Jon waits on tables and is paid $5.00 an hour. In a 40-hour period, he earned $210.50 in tips. What were his total earning
Nataly [62]

Add together the income from straight pay and that from tips:

($5/hr)(40 hrs) + $210.50

$200 + $210.50 = $410.50.


Jon's total earnings for the week were $410.50.

7 0
4 years ago
A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
GREYUIT [131]

Answer:

(a) Null Hypothesis, H_0 : \sigma = 10 beats per minute  

     Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute  

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute.

Let \sigma = <u><em>population standard deviation for the pulse rates of men</em></u>.

(a) So, Null Hypothesis, H_0 : \sigma = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is <u>One-sample chi-square test</u> for standard deviation;

                             T.S.  =  \frac{(n-1)\times s^{2} }{\sigma^{2} }  ~  \chi^{2}__n_-_1  

where, s = sample standard deviation = 10.1 beats per minute

            n = sample of men = 36

So, <u><em>the test statistics</em></u> =  \frac{(36-1)\times 10.1^{2} }{10^{2} }  ~ \chi^{2}__3_5

                                   =  35.70  4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

                    P-value = P(\chi^{2}__3_5 > 35.704) = <u>0.4360</u>

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

3 0
4 years ago
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miss Akunina [59]

Answer:

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Step-by-step explanation:

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