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Dafna1 [17]
4 years ago
5

Julio is paid 1.4 times his normal hourly rate for each hour he works over 30 hours in a week. Last week he worked 35 hours and

earned $436.60. Write and solve an equation to find Julio's normal hourly rate, r. Explain how you know that your answer is reasonable.
Mathematics
2 answers:
irina1246 [14]4 years ago
8 0

Answer:

Normal earning rate of Julio is $11.78

Step-by-step explanation:

Let the normal hourly rate of earning = $r per hour

Then for overtime earnings = 1.4 × $r per hour

Julio works for 30 hours so earnings = 30(r)

last he did overtime for 5 hours (35 - 30) so the earning for the overtime = 5(1.4r) = 7r

Total earning of Julio = 30r + 7r = $436

or 37r = $436

r = \frac{436}{37} = $11.78

Therefore, normal rate of Julio's earning is $11.78.

patriot [66]4 years ago
6 0
N = regular rate
1.4n= overtime rate

30n + 5(1.4n) = $436.60  (30 hours x regular rate n)  plus 5 hours at 1.4 times reg rate.

30n + 7n = $436.60    now add like terms
37n = $436.60    now divide both sides by 37
n= $11.80   
Check to see if my answer is reasonable 30 x 12= 360  + 5 x 10 x 1.4 = about 70 (notice that I used 10 instead of 12 here so I could do it in my head)
360 + 70 = 430 - close enough - my answer is reasonable
Check:
(30 x $11.80) + 5 ( 1.4 x 11.80
354 + 5( 16.52)
354 + 82.60 = 436.60    ($436.60)
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I. In a shipment of 300 connecting rods, the mean tensile strength is found to be 45 kpsi and has a standard deviation of 5 kpsi
CaHeK987 [17]

Answer:

a) Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

b) 260 rods are expected to have a strength between 39.4 and 60 kpsi

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 45, \sigma = 5

(a) Assuming a normal distribution, how many rods can be expected to have a strength less than 39.4 kpsi?

The percentage of rods with a stength less than 39.4 is the pvalue of Z when X = 39.4. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.4 - 45}{5}

Z = -1.12

Z = -1.12 has a pvalue of 0.1314

13.14% of rods have a strength less than 39.4 kpsi.

Out of 300

0.1314*300 = 39.42

Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

(b) How many are expected to have a strength between 39.4 and 60 kpsi?

The percentage of rods with a stength in this interval is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 39.4. So

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 45}{5}

Z = 3

Z = 3 has a pvalue of 0.9987

X = 39.4

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.4 - 45}{5}

Z = -1.12

Z = -1.12 has a pvalue of 0.1314

0.9987 - 0.1314 = 0.8673

86.73% of the rods are expected to have a strength between 39.4 and 60 kpsi

Out of 300

0.8673*300 = 260

260 rods are expected to have a strength between 39.4 and 60 kpsi

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4 years ago
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3 years ago
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Answer:

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Step-by-step explanation:

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