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Kobotan [32]
3 years ago
13

Talia gave her hairdresser a 20 percent tip, which amounted to $7. What was the price of the service before the tip?

Mathematics
2 answers:
kondor19780726 [428]3 years ago
8 0
<h2>Answer:</h2>

The price of the service that was charged before the tip was:

                            $ 35

<h2>Step-by-step explanation:</h2>

Let the price of the service be $ x.

Now it is given that:

Talia gave her hairdresser a 20 percent tip, which amounted to $7.

This means that the 20% of total service charge=7

i.e.  20% ×x =7

i.e. 0.20×x=7

i.e. x=7/0.20

i.e. x=35

Hence, the price of the service that was paid before the tip is: $ 35

mina [271]3 years ago
4 0
I don't know if I did it right but here's my guess. I think you multiply 7 by 20% and that gives you 1.40 as a result. Compared to the other method I used (division), I'm pretty sure I'm right since the other result was 35. Doesn't make sense if the end amount was $7.
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Answer:

<h2>x=-5</h2>

Step-by-step explanation:

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2 years ago
A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than
vovikov84 [41]

Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

H_0: p \leq 0.05

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

H_a: p > 0.05

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that \mu = 0.05, \sigma = \sqrt{0.05*0.95}

A random sample of 300 items taken from the production line contained 27 defective items.

This means that n = 300, X = \frac{27}{300} = 0.09

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}

z = 3.18

Pvalue of the test:

Testing if the mean is greater than a value, which means that the pvalue of the test is 1 subtracted by the pvalue of Z = 3.18, which is the probability of a finding a sample proportion of 0.09 or higher.

Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

5 0
3 years ago
Can someone help, please?
Kipish [7]
I think the answer is c
5 0
2 years ago
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