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Liula [17]
3 years ago
6

What is the length of DE

Mathematics
1 answer:
Tems11 [23]3 years ago
5 0

nce BE=EC, BC = 2BE. According the principle of proportionality of triangle,BE/DE=BC/AC.

This can be re-written as:

BE/DE= 2BE/ AC

DE= AC * (BE/2BE)

DE= AC * (0.5)

Since we do not know the value of AC from statement (2) therefore not sufficient. However substituting the value of AC from statement (1), we get DE= 14 * 0.5 = 7

Therefore answer is C, both statements together are sufficient but neither alone is sufficient.

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Solve the question 3/4 times 8
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6.24

Step-by-step explanation:

3/4=.75

.75*8=6.24

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T costs $6 per pound for peanuts at Blessed Food Market. Which of the following represents the range of the function in terms of
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"Cost of Peanuts Purchased" My Guy.

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Ab+c=de-f solve for d
lisabon 2012 [21]

Answer:

d=(ab+c+f)/e

Step-by-step explanation:

ab+c=de-f

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3 years ago
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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.

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3 years ago
What is the range of this data set?
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