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3 years ago

Find the value of two numbers if their sum is 130 and their difference is 28

Mathematics

2 answers:

Naddik [55]3 years ago

7 0

Answer:

Step-by-step explanation:

let the two numbers are x and y

x+y =130

x-y=28

on adding both the above equations

x+y+x-y = 130+28

2x = 158

x = 79

putting the value of x in any of the equation

79+y = 130

y = 130 -79

y = 51

Leto [7]3 years ago

3 0

79 and 51

They add to get 130 and subtract to get 28

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<h3>Probability</h3>

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Answer:

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Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they choose the leading brand, or they do not. The probability of a customer choosing the leading brand is independent of any other customer, which means that the binomial probability distribution is used to solve this question.

Binomial probability 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.

The leading brand of dishwasher detergent has a 30% market share.

This means that $p = 0.3$

A sample of 25 dishwasher detergent customers was taken.

This means that $n = 25$

a. What is the probability that 10 or fewer customers choose the leading brand?

This is:

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 0) = C_{25,0}.(0.3)^{0}.(0.7)^{25} = 0.0001$

$P(X = 1) = C_{25,1}.(0.3)^{1}.(0.7)^{24} = 0.0014$

$P(X = 2) = C_{25,2}.(0.3)^{2}.(0.7)^{23} = 0.0074$

$P(X = 3) = C_{25,3}.(0.3)^{3}.(0.7)^{22} = 0.0243$

$P(X = 4) = C_{25,4}.(0.3)^{4}.(0.7)^{21} = 0.0572$

$P(X = 5) = C_{25,5}.(0.3)^{5}.(0.7)^{20} = 0.1030$

$P(X = 6) = C_{25,6}.(0.3)^{6}.(0.7)^{19} = 0.1472$

$P(X = 7) = C_{25,7}.(0.3)^{7}.(0.7)^{18} = 0.1712$

$P(X = 8) = C_{25,8}.(0.3)^{8}.(0.7)^{17} = 0.1651$

$P(X = 9) = C_{25,9}.(0.3)^{9}.(0.7)^{16} = 0.1336$

$P(X = 10) = C_{25,10}.(0.3)^{10}.(0.7)^{15} = 0.0916$

Then

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0001 + 0.0014 + 0.0074 + 0.0243 + 0.0572 + 0.1030 + 0.1472 + 0.1712 + 0.1651 + 0.1336 + 0.0916 = 0.9021$

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand

3 0

3 years ago

Find the value of two numbers if their sum is 130 and their difference is 28​

Find the value of two numbers if their sum is 130 and their difference is 28