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3 years ago

The sum of 3 consecutive even integers is 78. What is the largest of the integers?

Mathematics

2 answers:

Vsevolod [243]3 years ago

8 0

Answer:

Step-by-step explanation:

Let the first even integer be a.

The second even integer = a + 2

The third even integer = 2 + (a + 2) = a + 4

The sum of three consecutive even integers = 78

78 = a + (a + 2) + (a + 4)

78 = 3a + 6

72 = 3a

a = 24

The largest even integers is the third even integer = a + 4 = 24 + 4 = 28

Maksim231197 [3]3 years ago

3 0

Answer:

is the sum 78 or 7,878 (i.e. was there a stutter)?

In any event, this sum must be even, so 3 consecutive even integers would be n, n+2 & n+4 whose sum is 3n + 6.

Set that to either 78 or 7878, whichever is intended, & solve for n. If 78 is their grand total, then those 3 consecutive even numbers have to be 24, 26 & 28.

No stress! Just logic! Keep the faith!

Step-by-step explanation:

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b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

$M = \frac{a + b}{2}$

The standard deviation of the uniform distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

Uniform distribution between 2.93 and 3.1 volts

This means that $a = 2.93, b = 3.1$ . So

Mean:

$M = \frac{2.93 + 3.1}{2} = 3.015$

Standard deviation:

$S = \sqrt{\frac{(3.1 - 2.93)^{2}}{12}} = 0.049$

What is the probability that a battery has a voltage less than 2.98?

$P(X \leq 2.98) = \frac{2.98 - 2.93}{3.1 - 2.93} = 0.2941$

So the correct answer is:

b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

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