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Nady [450]
2 years ago
13

Margit works in a factory. She receives a salary of $12 per hour and piecework pay of 12 cents per unit produced. Last week she

worked 38 hours and produced 755 units.
a. What was her piecework pay?

b. What was her total hourly pay for the week?

c. What was her total pay for the week?

d. What would her total weekly salary have been if she produced 0 units?
Mathematics
1 answer:
lilavasa [31]2 years ago
4 0

Answer:

d is 456

a is 90.60

b is 14.38

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

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3 years ago
Write 2 4/11 as a decimal
Yanka [14]

2.36 repeating                                    

THIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIISSSSSSSSSSSSSSSSSS IS THE ANSWER!

7 0
3 years ago
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Bob wants to put down new sod in his backyard except for the part set aside for his flower garden the diagram shows bobs backyar
zmey [24]

Answer:

Bob will need 235\ yd^{2} of sod

Step-by-step explanation:

step 1

Find the area of the backyard

A=20(14)=280\ yd^{2}

step 2

Find the area of the flower garden

A=5(9)=45\ yd^{2}

step 3

Find the difference

280\ yd^{2}-45\ yd^{2}=235\ yd^{2}

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3 years ago
Christine has a rope that is 9 meters long. She cuts away a plece that is 3.29 meters long. How long is the remaining plece of r
natima [27]

Answer:

9.00-3.29= 5.71 meters rope remaining

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3 years ago
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a) The weekly wages of employees of Volta gold are normally distributed about a mean of $1,250 with a standard deviation of $120
Fantom [35]

Answer:

0.7102

0.8943

0.3696

Step - by - Step Explanation :

A.) Between $1320 and $970

P(Z < 1300) - P(Z < 970)

find the Zscore of each scores and their corresponding probability uinag the standard distribution table :

P(Z < (x - μ) /σ) - P(Z < (x - μ) / σ))

P(Z < (1320 - 1250) /120) - P(Z < (970 - 1250) / 120))

P(Z < 0.5833) - P(Z < - 2.333)

0.7200 - 0.0098 = 0.7102 (Standard

=0.7102

B.)Under 1400

x = 1400

P(Z < 400)

P(Z < (x - μ) /σ)

P(Z < (1400 - 1250) /120)

P(Z < 1.25) = 0.8943

C.) Over 1290

P(Z > 1290)

P(Z < (x - μ) /σ)

P(Z > (1290 - 1250) /120 = 0.3333

P(Z > z) = 1 - P(Z < 0.3333) = P(Z < 0.3333) = 0.6304

P(Z > 0.3333) = 1 - 0.6304 = 0.3696

4 0
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