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kenny6666 [7]
3 years ago
13

Gianna is paid $90 for 5 hours of work. • How much money does Gianna make per hour?

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

18 dollars per hour

Step-by-step explanation:

Oliga [24]3 years ago
4 0

Answer:

18 dollars a hour.

Step-by-step explanation:

all you have to do is divide 90 from 5

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Please I need help and thx<br>please show how you did it
rosijanka [135]

Answer:

1.) 9.2

2.)

625

633

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8.81

Step-by-step explanation:

I'm gonna assume that cm= compounded monthly

1.)

effective rate: .153/12= .01275

x= payments

2590.67=300*\frac{1-(1+.01275)^{-x}}{.01275}\\.110103476=1-1.01275^{-x}\\.889896525=1.01275^{-x}\\\log_{1.01275}.889896525=-x\\x=9.207

2.)

If there is no interest rate attached to financing through the deal the payment is just

37500/60 = 625

The monthly payment from the bank has a present value of 37500-3000=34500

and the effective rate is .039/12= .00325

34500=x\frac{1-(1.00325)^{-60}}{.00325}\\34500=54.43234738x\\x=633.81

Finally, the amount we save is just the difference

633.81-625=8.81

6 0
3 years ago
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The height of a cylinder is 8 centimeters. The circumference of the base of the cylinder is 14 pi. What is the measurement of th
timurjin [86]

Answer:

15

Step-by-step explanation:

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2 years ago
The mean preparation fee H&amp;R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

  • Me is the margin of error from the mean
  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

  • N is the sample size
  • z is the corresponding z-score in 95% probability (1.96)
  • s is the standard deviation (50)
  • ME is the margin of error (8)

then N≥(\frac{1.96*50}{8} )^2 ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

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3 years ago
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