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3 years ago

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21. Susan traveled 114 miles in 2 hours. If she keeps going at the same rate, how long will it take her to go the remaining 285

miles of her trip?
a. 5 hours
b. 3 hours
c. 7 hours
d. 4 hours

1 answer:

sveticcg [70]3 years ago

5 0

Answer:

5 hours are the answer long will take her to go the remaining

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Find the slope of the line containing the pair of points.<br> (6,5) and (0,0)

GarryVolchara [31]

Answer: 5/6

Step-by-step explanation:

Use slope formula of y2-y1/x2-x1. So here it doesn’t matter which is which so I did (6,5) x1 and y1 and (0,0) x2 and y2

So plugging it in it would be 0-5/0-6 = -5/-6 which simplifies to 5/6

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1 year ago

FREE BRAINLY TO THE CORRECT ANSWER!!

Lyrx [107]

Answer: -3

Step-by-step explanation:

y = mx + b (slope intercept formula)

9x - 3y = -54

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3 years 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:

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

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago

Help help help help help

SIZIF [17.4K]

Answer:

1) 2b + 6

2) 8w - 12

3) 2b + 6 + 6x + 20

4) 9w^2 + 27 + 8 - 27

5) Sorry.

Step-by-step explanation:

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3 years ago

Help I need help please

s344n2d4d5 [400]

a) Negative correlation

Negative correlation corresponds to points that move down as you go from left to right on the scatter plot.

b) Positive correlation

Positive correlation corresponds to points that move up as you go from left to right on the scatter plot.

c) No correlation

Scattered points on the scatter plot make no relationship.

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3 years ago