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3 years ago

13

the endpoints of segment LE are L(-2,2) and F(3,1). the endpoints of segment JR are J(1,-1) and R(2,-3). what is the approximate

difference in the lengths of the two segments?

2 answers:

Alenkinab [10]3 years ago

7 0

Just think a bit don’t procrastinate! You got this

Snezhnost [94]3 years ago

7 0

S=~ 5.099 is the awnser

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How to find the measure of 2 supplementary angles if the differences in the measures of the angles is 102 degrees?

uysha [10]

You would do 180 - 102 to find the answer

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2 years ago

Michael canned 4,713 quarts of blueberries over the summer. By the end of the year he had used 1/3 of what he had canned. How ma

aleksandr82 [10.1K]

Answer:

he used 1,571

Step-by-step explanation:

if you divide 4,713 by 3, you get 1,571 and that amount is equal to just 1/3 of the amount.

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3 years ago

Suppose the average tread-life of a certain brand of tire is 42,000 miles and that this mileage follows the exponential probabil

Ahat [919]

Answer: The probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

Step-by-step explanation:

The cumulative distribution function for exponential distribution is :-

$P(X\leq x)=1-e^{\frac{-x}{\lambda}}$ , where $\lambda$ is the mean of the distribution.

As per given , we have

Average tread-life of a certain brand of tire : $\lambda=\text{42,000 miles }$

Now , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles will be :

$P(X\leq 65000)=1-e^{\frac{-65000}{42000}}\\\\=1-e^{-1.54761}\\\\=1-0.212755853158\\\\=0.787244146842\approx0.7872$

Hence , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

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3 years ago

Which expression is equivalent to?

guapka [62]

Answer: $\frac{2x\sqrt[4]{y^{2}}}{3}$

Step-by-step explanation:

$\sqrt[4]{\frac{16}[81}} \sqrt[4]{\frac{x^{11}y^{8}}{x^{7}y^{6}}}\\\\=\frac{2}{3} \sqrt[4]{x^{4}y^{2}}\\\\=\frac{2x\sqrt[4]{y^{2}}}{3}$

4 0

2 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744$

99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

7 0

3 years ago