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3 years ago

(Number 8 answer ?) I’m having trouble with this concept.

Mathematics

1 answer:

coldgirl [10]3 years ago

5 0

Answer: C

I put the working in the other question you put up, but to put it briefly, multiply everything by 6 because all the options provided have a -6 inside that was meant to be the -1

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Divide twice the variable p by the product of 5 and q

Law Incorporation [45]

Answer:

Step-by-step explanation:Solution :

Given variable = p

According to the problem given ,

Twice the variable p divided by the

product of 5 and q

= ( 2p )/ ( 5q )

= 2p/5q

5 0

2 years ago

let's one two and three have the following relationships <1and <2 are adjacent right angles <1 and <3 are vertical a

Fed [463]

Answer:

180 Vertical angles are opposite angles that share only a vertex. Since ∠3 is adjacent to both ∠1 and ∠2, this means that ∠3 shares a side and vertex with both of these angles.

This means that ∠3 and ∠1 form a straight line; this makes them a linear pair, which makes their sum 180°.

8 0

2 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

2 years ago

Sonya is 22 years younger than her mother. the sum of their age is 32. how old are they

Fantom [35]

Answer:

Sonya is ten her mother is 22

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the greatest common factor of the terms in the polynomial 4x^4-32x^3-60x^2?

jonny [76]

Answer:

$\mathrm{Factor}\:4x^4-32x^3-60x^2:\quad 4x^2\left(x^2-8x-15\right)\\$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^3=xx^2\\x^4=x^2x^2\\=4x^2x^2-32xx^2-60x^2\\\mathrm{Rewrite\:}60\mathrm{\:as\:}4\cdot \:15\\\mathrm{Rewrite\:}32\mathrm{\:as\:}4\cdot \:8\\=4x^2x^2-4\cdot \:8xx^2-4\cdot \:15x^2\\\mathrm{Factor\:out\:common\:term\:}4x^2\\=4x^2\left(x^2-8x-15\right)$

<em>Hope this helps and have a great day!!!</em>

<em> $Sofia$ </em>

4 0

2 years ago