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3 years ago

G=4cm-3m solve for m

Mathematics

1 answer:

adell [148]3 years ago

6 0

m • (g4c - 3)

(1):g4 was replaced by g^4.

Pulling out like terms:

2.1:Pull out like factors:

g4cm - 3m = m • (g4c - 3)

Trying to factor as a Difference of Squares :

2.2:Factoring: g4c - 3

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Answer:

14 children, 5 women, 1 man

Step-by-step explanation:

14 children which gives us 7 loves. 1 man which gives us 3 loves. 5 women which gives us 10 loves.

People: 14 + 1 + 5 = 20

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3 years ago

Is it ever possible that cos (A−B)=cos ⁡A−cos ⁡B? Why

svetlana [45]

Answer:

yes, there are an infinite number of solutions

Step-by-step explanation:

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3 years ago

A die is thrown. What is the probability of getting 6? (a) 0 b 1/6 c 1/2 d 1

Firdavs [7]

Answer:

1/6

Step-by-step explanation:

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3 years ago

Plz help it’s due tonight!!!

jasenka [17]

sorry

Step-by-step explanation:

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3 years ago

Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car i

Vesna [10]

Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

Step-by-step explanation:

Let X = r = a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = p = 0.35.

The sample selected is of size n = 12.

The random variable X follows a Binomial distribution with parameters n = 12 and p = 0.35.

The probability function of a binomial random variable is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x}$

(a)

For X = 0 the probability is:

$P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006$

For X = 1 the probability is:

$P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037$

For X = 2 the probability is:

$P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109$

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

The expected value of a Binomial distribution is:

$E(X)=np$

The expected number of vehicles out of 12 that will tailgate is:

$E(X)=np=12\times0.35=4.2$

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

$SD(X)=np(1-p)$

The standard deviation of vehicles out of 12 that will tailgate is:

$SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\$

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.

4 0

3 years ago