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Alika [10]
3 years ago
15

Please help me, worth LOTS of points

Mathematics
1 answer:
Aliun [14]3 years ago
4 0

Answer:

a) 91\,+\,0.14 \,x \leq 140   where "x" stands for the number of miles driven.

b) He can drive as far as 250 miles to keep the rental cost limited to $140.

Step-by-step explanation:

a) Robert wants to make sure that the addition of the costs coming from the car rental per week ($91) plus the amount paid for the coverage of "x" number of miles (which goes as $0.14 times x) does not exceed $140 (which is the same as saying that this total cost must be smaller than or equal to $140.

In math terms, such is written as:

91\,+\,0.14 \,x \leq 140

where "x" stands for the number of miles driven.

b) the total number of miles (x) he is allowed to cover given the $140 restriction is obtained by solving for "x" (the number of driven miles) in the inequality of part a):

91\,+\,0.14 \,x \leq 140\\0.14\,x\leq 140-91\\0.14\,x\leq 49\\x\leq \frac{49}{0.14} \\x\leq 350

which tells us that the number of driven miles (x) has to be smaller or equal to 350 miles. Then he can drive as far as 250 miles to keep the rental cost limited to $140.

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A = ½ (10 + 12) (6)

A = 66

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
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Answer:

1.2%

Step-by-step explanation:

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So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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Answer: From least to greatest: -2.4, -0.8, 0.2, 0.9, 1.6. Hope this helps you! :D

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