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3 years ago

the sum of two numbers is 78. one of the numbers is 4 more than the other number, what are the two numbers?

Mathematics

2 answers:

ladessa [460]3 years ago

7 0

The two numbers are 41 & 37.

The way to solve this is by setting up an equation. We know that 2 numbers are going to equal 78, so here would be the general equation (let x represent the number we don't know):

(x) + (x + 4) = 78

Now, we need to simplify it (combine like terms):

2x + 4 = 78

Then you can solve it like any other equation. Subtract 4 on both sides:

2x = 74

Divide by 2 on both sides:

x = 37.

So, now you know one number is equal to 37 and that the other is 37 + 4.

That's how you get 37 & 41.

lilavasa [31]3 years ago

7 0

X = ?

X +4= ?

(X+4) + x = 78

2X + 4 = 78. Subtract 4 from both sides

You get 2X = 74. Divide both sides by 2 and you get x= 37

Now plug 37 in and you get 37 and 41

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3 years ago

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst

grandymaker [24]

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(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let X = number of telephone calls.

The average number of calls per minute is, λ = 3.0.

The random variable X follows a Poisson distribution with parameter λ = 3.0.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...$

(a)

Compute the probability of X = 0 as follows:

$P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498$

Thus, the probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of X > 5 as follows:

P (X > 5) = 1 - P (X ≤ 5)

$=1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084$

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of X = 3 as follows:

$P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09$

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of X ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

$=\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578$

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

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3 years ago