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3 years ago

The sum of 5 consecutive integers is 110, what is the fourth number in this sequence?

Mathematics

1 answer:

Xelga [282]3 years ago

5 0

The answer is: " 23 " .

→ The fourth number in the sequence is: " 23 " .

_____________________________________

Explanation:

To solve:

" x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 110 " ;

in which:

"x" = The first number in these sequence;

"(x + 1 )" = the second number in the sequence;

"(x + 2)" = the third number in the sequence;

"(x + 3)" = the fourth number in the sequence;

"(x + 4)" = the fifth number in the sequence;

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Solve for the "fourth number in the sequence" ; or: "(x + 3)" ;

_____________________________________

Given: " x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 110 " ;

↔ " x + x + 1 + x + 2 + x + 3 + x + 4 = 110 " ;

→ Solve for "x" ;

→ then, solve for "(x + 3)" ;

→ which is the fourth number in this sequence;

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→ " x + x + 1 + x + 2 + x + 3 + x + 4= 110 " ;

"5x + 1 + 2 + 3 + 4= 110 " ;

" 5x + 10 = 110 " ;

Solve for "x" :

Subtract "10" from EACH SIDE of the equation; as follows:

" 5x + 10 - 10 = 110 - 10 " ;

to get :

" 5x = 100 " ;

Now, divide EACH SIDE of the equation by "5" ;

to isolate "x" on one side of the equation; & to solve for "x" ; as follows:

5x / 5 = 100 / 5 ;

to get:

" x = 20 " .

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Now, to find the fourth number in the sequence:

→ " (x + 3) " ;

→ Substitute "20" for "x" ;

" x + 3 = 20 + 3 = 23" ;

_____________________________________

The answer is: " 23 " .

The fourth number in the sequence is: " 23 " .

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Let us check our work:

If there are five "5" numbers in the sequence, and "23" is the "fourth number" , then: "24" is the "fifth number" .

As such: "22" is the "third number" ; "21" is the "second number" ; and "20" is the "first number" . Is this consistent with: "x = 20" as the "first number" ? Yes!

Thus, "20 + 21 + 22 + 23 + 25 = ? 110 ? ? ;

→ 20 + 21 = 41 ;

→ 41 + 22 = 63 ;

→ 63 + 23 = 86 ;

→ 86 + 24 = 110 . Yes!

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Hope this answer is helpful!

Best wishes in your academic pursuits—

and within the "Brainly" community!

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2 years ago

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prisoha [69]

#2
2x - 4y = -2
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------------------subtract
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check:

2(-5)- 4(-2) = -2
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2 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

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