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3 years ago

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Josh Wilson is traveling to California from South Carolina for summer vacation. The states are approximately 2500 miles apart dr

iving cross-country. He stops at his parent's house in Oldham County, Texas, which is (12x 106) miles away from South Carolina and (5x + 685) miles away from California. Determine if Josh's parent's house in Texas the midpoint between South Carolina and California? Justify your answer.

2 answers:

grandymaker [24]3 years ago

8 0

Answer:

Yes, Josh's parent's house in Texas is the midpoint between South Carolina and California

Step-by-step explanation:

Let S represents South carolina C represents California and T represents Texas,

Given,

ST = 12x - 106,

TC = 5x + 685,

Also, SC = 2500 miles,

Suppose T is the midpoint of S and C,

⇒ ST = TC

12x - 106 = 5x + 685

7x = 791

⇒ x = $\frac{791}{7}$ = 113

So, ST = $12(113)+106 =1356 - 106 = 1250$

Now,

$2\times ST = 2\times 1250=2500$

⇒ 2 ST = SC

Hence, T is the midpoint of S and C.

MakcuM [25]3 years ago

5 0

Given, distance between South Carolina and Texas = $12x-106$

Distance between Texas and California = $5x+685$

Let's assume, both the distance is same.

$12x-106=5x+685$

=> $12x-5x = 685+106$

=> $7x = 791$

=> $x=113$

Plug x = 113 in $5x+685$

$5(113)+685 = 1250$ Miles.

Similarly, $12(113)-106=1250$ miles

Since 2500 divided by 2 = 1250 miles, hence our assumption was correct, Texas lies between both the cities.

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Let x,y, and z are the numbers of two-point field goals, numbers of three-point field goals, and the number of free-throws (one-point goal) respectively.

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Answer:

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And rounded up we have that n=2401

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

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Part b

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And rounded up we have that n=626

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