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2 years ago

A printer can print 40 pages in 1.6 minutes

1 answer:

4 0

Given:
40 pages in 1.6 minutes

40 / 1.6 = 25 pages per minute

25 x 5 minutes = 125 pages

40 / 1.6 = x / 5
40*5 = 1.6x
200 = 1.6x
200/1.6 = x
125 = x pages

25 : 1 = 150 : x
25x = 150
x = 150/25
x = 6 minutes

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If f(x) =ax+b ,f(-3)=-4 and f(3)=2, find the value of a and b please help

hichkok12 [17]

Answer:

a = 1

b = -1

Step-by-step explanation:

First, let's make equations out of those given f(x)'s

-4 = -3a + b

2 = 3a + b

We can use elimination method to get rid of the a real quick

-4 = b

2 = b

Add those together

-2 = 2b

Divide both sides by 2

-2/2 = 2b/2

b = -1

Plug in the new b in one of the original equations; I'm gonna use the bottom one

2 = 3a - 1

+ 1 + 1

3 = 3a

3/3 = 3a/3

a = 1

4 0

1 year ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$