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Nadusha1986 [10]
4 years ago
11

Least common factor of 60 and 72

Mathematics
1 answer:
Arlecino [84]4 years ago
8 0
The Least Common Factor of 60 and 72 is 360
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Tyje's paycheck varies directly with the number of hours he works. If he makes $238 for working 28 hours:
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Answer:

238/28=$8.50 a hour for 28 hours

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4 years ago
A. What is the circumference of the
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Answer:

Step-by-step explanation:

a. The diameter of the circle will be 30 cm

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b. Required percentage = (94.2) * 100 / (4*30)

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A music teacher has 4 drum kits. Each kit has 2 drumsticks. Each drumsticks costs $3 how many drumsticks does she have what is t
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A music teacher has 4 drum kits. Each kit has 2 drumsticks. Each drumsticks costs $3 how many drumsticks does she have what is the cost to replace them all?

answer: she will have 8 sticks and $24 to replace

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4 0
3 years ago
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Minchanka [31]
<h2><u>Answer with explanation</u>:</h2>

Let \mu be the distance traveled by deluxe tire .

As per given , we have

Null hypothesis : H_0 : \mu \geq50000

Alternative hypothesis : H_a : \mu

Since H_a is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.

Test statistic : z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

where, n= sample size

\overline{x}= sample mean

\mu= Population mean

s=sample standard deviation

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z=\dfrac{46800-50000}{\dfrac{8000}{\sqrt{31}}}=-2.23

By using z-value table,

P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23)   [∵P(Z≤-z)=1-P(Z≤z)]

=1-0.9871=0.0129

 Decision : Since p value (0.0129) < significance level  (0.05), so we reject the null hypothesis .

[We reject the null hypothesis when p-value is less than the significance level .]

Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.

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3 years ago
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