Which equation represents the graphed function?

Solve by factorising the equation 8x^2-30x-27=0

IrinaK [193]

I got 9/2,-3/4. hope it helps

6 0

3 years ago

Tom has three times as many pencils as pens but has a total of 100 writing utensils. How many pencils does Tom have? How many pe

Genrish500 [490]

Answer:

There are 75 pencils and 25 pens.

Step-by-step explanation:

If Tom has three times as many pencils than pens then,

Pencils = 3(Pens)

If there are a total of 100 writing utensils then,

Pencils + Pens = 100

Substitute the word "pencils" in this equation to 3(Pens),

3(Pens) + Pens = 100

Now remember there is almost a 'hidden' one beside the word "Pens" so,

3Pens + 1Pens = 100

Add the two together,

4Pens = 100

Divide,

Pens = 25

Substitute 25 into the first equation,

Pencils = 3(25)

Pencils = 75

Thus, there are 75 pencils and 25 pens.

3 0

3 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 $n = 4$