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3 years ago

8

A middle school is selling candy bars to raise money for new calculators. The graph shows the number of candy bars sold, x, and

the amount of money collected, y.
Enter a value that completes the equation correctly.

Y=__x

2 answers:

Aliun [14]3 years ago

6 0

Y equals 2.5x because it increases 2.5

Naily [24]3 years ago

4 0

Y=2.5x as each time it increases by 2.5

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What’s -3 1/9 — (-5 5/9)

Svetradugi [14.3K]

-28/9-(-5 5/9)
-28/9-(50/9)
-28/9-50/9
22/9
= 2 4/9= 2.444444
hope this helps!

7 0

4 years ago

Read 2 more answers

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

Is the relation shown in the arrow diagram a function?

Irina-Kira [14]

Answer:

Yes. Every unique input has a unique output.

Step-by-step explanation:

Since every input has a different output, therefore it's a function. It's not a function when the input have more than 1 output. When graphing, make sure you take the vertical line test to see whether or not the graph is a function.

4 0

3 years ago

Help me pleaseeeeeeeeee

gayaneshka [121]

THE ANSWER IS A.....

7 0

4 years ago

If figure 5 has 12 dots how many dots does figure 50 have

Katarina [22]

The answer is 600 dots

7 0

3 years ago