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Tcecarenko [31]
2 years ago
7

This week, I need you to put in some overtime,which will require you to work 15% longer than your normal 40 hours per work sched

ule. Then I should plan on working a total of how many hours this week?
Mathematics
1 answer:
fiasKO [112]2 years ago
4 0

Answer: 46 hours

Step-by-step explanation:

15% of the normal working hours is;

\frac{15}{100} × 40 = 6 hours

∴ 15% longer for this week will mean,

6 hours + 40 hours = 46 hours

so, plan on working a total of 46 hours this week

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Quadrilateral WXYZ is on a coordinate plane. Segment XY is on the line x − 3y = −12, and segment WZ is on the x − 3y = −6. Which
Nostrana [21]

Answer:

Equation shown

Step-by-step explanation:

Eq(1): x-3y=-12

X=-12+3y

=-4+y

Eq(2): x-3y=-6

X=-6+3y

=-2+y

4 0
2 years ago
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The median number of cars sold by 11 sales representatives in a certain month is 6. The range of a number of cars sold by those
Luda [366]
It is given that the range of the number of cars sold is 4. Therefore if the fewest cars sold is 3, then the greatest number of cars sold will be 3 + 4 = 7.
The median number of cars sold is 6, so it is possible that the greatest number of cars sold is 7.
The statement is TRUE.
5 0
2 years ago
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Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

6 0
2 years ago
NEED HELP!
TEA [102]

Answer:

30 minutes

Step-by-step explanation:

20 * 96 / 64 = 30

8 0
2 years ago
If your front lawn is 17.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snowflakes every minu
Alexus [3.1K]

Answer: 42.84 kg

Step-by-step explanation: Lawn is 17 ft x 20 ft.

Its area is S = 17 x 20 = 340 ft²

Each ft² of lawn accumulates 1050 snowflakes/min, 1ft² = 1050 s.f./min.

As 1 hour has 60 min, the lawn accumulates 63000/hour →

1050 x 60 = 63000

1ft² = 63000s.f./hour

This way, 340 ft² of lawn will accumulate 21,420,000 s.f/hour →

340 x 63000 = 21420000

As 1 snowflake has 2mg,  21,420,000 will have 42,840,000 mg.

As 1kg = 1,000,000 mg

42,840,000/1,000,000 = 42.84 kg

3 0
2 years ago
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