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zhenek [66]
2 years ago
9

Juan makes $9 per hour at his job. The equation represents how much money he made yesterday. T=9h if Juan make (T) $76. 50 yeste

rday, how many hours did he work?
Mathematics
1 answer:
Roman55 [17]2 years ago
6 0
8.5 hours


T = 9 • h (hours worked)

$76.50 = 9 • h

76.50/9 = h

h = 8.5 hours
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vivado [14]

Answer:

x=-4

Step-by-step explanation:

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Step-by-step explanation:

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Help fast!! please.
alexira [117]

Answer:

Mean = 10

Variance = 30.89

Standard Deviation = 5.56

Step-by-step explanation:

Find the Mean, Variance and Standard deviation

6, 14, 8, 2, 11, 7, 12, 4, 18, 18

Step 1

We rearrange the data and find the Mean

2, 4, 6, 7, 8, 11, 12, 14, 18, 18

Mean = Sum of values/Number of values

= 2 + 4+ 6+ 7+ 8 + 11 + 12 + 14 + 18 +18/10

= 100/10

= 10

Step 2

We find the variance of this sample days ma

Variance = (x - Mean)²/n - 1

n = 10

Hence,

=( (2 - 10)² +(4 - 10)²+ (6 - 10)²+ (7 - 10)² + (8 - 10)² + (11 - 10)² + (12 - 10)² + (14 - 10)² + (18 - 10)² +(18 - 10)²) /10 -1

= ( 64 + 36 + 16 + 9 + 4 + 1 + 4 + 16 + 64 +64)/9

= 278/9

= 30.88888889

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6 0
3 years ago
Lamont has 30 books to pack in boxes. Each box holds 7 books. If he fills each
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5 0
3 years ago
A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is chang
zysi [14]

Answer:

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

Step-by-step explanation:

Let p1 be the sample of 1000 adults chosen in 2012 and

p2 be the sample of 1100 adults chosen in 2010

<u>Part a:</u>

The hypothesis that can be used to test for a significant difference between the population proportions for the two years is:

H0: p1-p2= 0           against the claim      Ha: p1-p2≠0

<u>Part b:</u>

The sample proportion indicating that their financial security was more that fair in 2012 is

p1^= 410/1000= 0.41

In 2010 is:

p2^= 385/1100= 0.35

<u>Part c:</u>

p^c= 410+385/1000+1100= 0.3785

q^c= 1-p^c= 1-0.3785= 0.6124

The critical region is z > ±1.96

The test statistic is

z= p1^- p2^ / sqrt ( p^cq^c( 1/n1+ 1/n2)

z= 0.41-0.35 /sqrt( 0.3785*0.6124(1/1000 +1/1100)

z=  0.06 / 0.021036

z= 2.85

Conclusion

The calculated z = 2.85 falls in the critical region z > 1.96 so we accept the null hypothesis that there is no difference between the population proportions for the two years.

P- value :  0.00466

The result is significant( accept the null hypothesis) for  value less than 0.05

<u>Part d:</u>

                                                     2012                         2010

Sample                                       1000                           1100

Proportions                               0.41                             0.35

Standard Error               √0.41( 0.59)/1000        √0.35( 0.65)/1100

                                               = 0.0155                0.01438

Standard Error for the difference = √0.0155² + 0.01438² =0.0211

p1-p2 ± z* standard error for the difference

0.41-0.35 ± 1.96 *( 0.0211)

0.06 ± 0.041356

-0.018644,   0.101356

The 95 % confidence interval estimate is (-0.018644,   0.10135

4 0
3 years ago
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