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Ludmilka [50]
3 years ago
8

Christy is paid time-and-half hours worked in excess for 40 hours and double time for hours worked on Sunday. If Christy had gro

ssed weekly wages of $592 for working 54 hours, 6 of which were on Sunday, what is her regular hourly rate?
Mathematics
1 answer:
Tamiku [17]3 years ago
7 0

Answer:

$9.25/hr

Step-by-step explanation:

Let regular hourly rate be "x"

Total hours worked = 54

6 of which are on Sunday (paid TWICE)

54 - 6 = 48

So, hours over 40 are:

48 - 40 = 8 hours (PAID 1.5 times)

So,

40 hours paid "x"

8 hours paid (1.5x)

6 hours paid (2x)

Which totals 592

We can write:

40x + 8(1.5x) + 6(2x) = 592

Now we can solve this for x to get our answer:

40x + 8(1.5x) + 6(2x) = 592\\40x+12x+12x=592\\64x=592\\x=\frac{592}{64}\\x=9.25

The regular hourly rate is $9.25/hr

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sweet [91]

The answer is A) x=0.3

Step-by-step explanation:

Move all terms containing xx to the left side of the equation.

14x−15.4=−11.2

Move all terms not containing x to the right side of the equation.

14x=4.214x=4.2

Divide each term by 14 and simplify

x=0.3

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Can someone help me I think I'm doing my math wrong
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Step-by-step explanation:

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3 years ago
What is the unit rate for 3 inches of rain in 6 hours?
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Answer:

18

Step-by-step explanation:

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4 0
3 years ago
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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
Describe the effect an increase in i, the interest rate applied to the present value, has on the monthly payment P in the formul
VARVARA [1.3K]

Answer:

B. An increase in i, the interest rate, will create an increase in P, the monthly payment.

Step-by-step explanation:

We have the formula for the monthly payment as,

P=\frac{i \times PV}{1-(1+i)^{-n} },

where P = monthly payment, i = rate of interest, PV = present value and n = time period.

Now, as i increase we get that (1+i) increases and so (1+i)^{n} increases.

This gives us that, \frac{1}{(1+i)^{n} } decreases and so 1-\frac{1}{(1+i)^{n}} decreases

Therefore, \frac{1}{1-(1+i)^{-n} } increases.

So, we get that as i increases , the value of P will increase.

Hence, option B is correct.

4 0
3 years ago
Read 2 more answers
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