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3 years ago

Also need help with this one

Mathematics

1 answer:

777dan777 [17]3 years ago

3 0

Answer:3/7+5/14

Step-by-step explanation:This is your answer because you have to have an addition problem, so you have to add them to get your addition sentence.

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The length of a rectangle is twice its width.

Furkat [3]

The perimeter is the sum of all a shapes sides.

so if length is two times the width, you could write it as 2w.

2w + 2w + w + w = 42. Simplified, it's 6w = 42.
Divide both sides by 6, and you get w = 7.

Since the length is 2w, 2 * 7 = 14, so the length is 14

Area of rectangle is length * width, so do 14 * 7 which is 98. Slap on some units and your answer is 98 sq m

7 0

3 years ago

Need help

Alika [10]

Answer:

Step-by-step explanation:

2*8= 16

4 0

3 years ago

Read 2 more answers

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

How to simplify 2mg x 5m²k

Damm [24]

hey, I'm not sure about this answer, but I think it's true, sorry

m(2g+5mk)

6 0

3 years ago

12e^5 divide by 3e^3

Paladinen [302]

The answer would be <em>29.5562243957</em>.

5 0

3 years ago