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Ipatiy [6.2K]
3 years ago
11

41/8 - 1/3 = what????​

Mathematics
2 answers:
Allisa [31]3 years ago
7 0

Answer:41/8 - 1/3= 115/24

Step-by-step explanation:

BartSMP [9]3 years ago
7 0
4 \frac{19}{24} . That should be the correct answer.
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An auto dealership sells minivans and sedans. In January, they sold 10 minivans and 20 sedans. In February, they sold 7 minivans
zaharov [31]

Answer:

january = 10/20 = 1/2

februry = 7/14 = 1/2

so, we conclude that the auto dealership didn't have a lower ratio, since the ratio is equal in both months.

<em>hope it helps :)</em>

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I need help showing my work for 67.42 \4
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A car rents for $250 per week plus $0.50 per mile.find the rental cost for a two-week trip of 500 miles for a group of three peo
Artyom0805 [142]
So 2 weeks mean
250x 2= 500
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3 years ago
A rectangular prism that has a length of 6 units, width of 4 units, and height of 5 units.
AlladinOne [14]

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Step-by-step explanation:

E D G E N U I T Y

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Brainliest?!?!

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3 years ago
Read 2 more answers
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

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

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

                               =  -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0
3 years ago
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