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Nikolay [14]
3 years ago
12

What is the Value of X?

Mathematics
2 answers:
marin [14]3 years ago
8 0
Solve is in the picture

<span>Good luck :)</span>

Yakvenalex [24]3 years ago
7 0
\dfrac{x+5}{6}=\dfrac{x+5+9}{12}\\\\\dfrac{x+5}{6}=\dfrac{x+14}{12}\ \ \ \ |cross\ multiply\\\\12(x+5)=6(x+14)\\\\12\cdot x+12\cdot5=6\cdot x+6\cdot14\\\\12x+60=6x+84\ \ \ |subtract\ 6x\ from\ both\ sides\\\\6x+60=84\ \ \ \ |subtract\ 60\ from\ both\ sides\\\\6x=24\ \ \ \ |divide\ both\ sides\ by\ 6\\\\\boxed{x=4}
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Answer: 2

Step-by-step explanation: 1+1=2

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The system of equations Y =-3x+2 and Y = 3x-6 is shown on the graph below.
Ira Lisetskai [31]

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A

Step-by-step explanation:

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3 years ago
The question is in the picture.
nordsb [41]
The first bullet is the answer
7 0
3 years ago
An instructor who taught two sections of engineering probability last term, the first with 20 students and thesecond with 30, de
Oliga [24]

Answer:

0.207

Step-by-step explanation:

This is an hypergeometric distribution problem

An hypergeometric distribution has the same sense as the discrete probabilities of binomial distribution, but unlike binomial distribution, hypergeometric distribution does not allow replacement.

Binomial distribution expresses the probability of picking k objects from n with replacement, but hypergeometric distribution expresses picking k objects from n without replacement, with the finite total population, N, containing K objects.

It is expressed mathematically as

h(k: n, K, N) = (ᴷCₖ)(ᴺ⁻ᴷCₙ₋ₖ)/(ᴺCₙ)

where

k = number of students in the 2nd section required to be in the first 15 graded projects (number of successes) = 10

n = total number of first graded projects (number of trials) = 15

K = number of students in the 2nd section of the class = 30

N = total number of students = 50

h(10: 15, 30, 50) = (³⁰C₁₀)(⁵⁰⁻³⁰C₁₅₋₁₀)/(⁵⁰C₁₅)

h(10: 15, 30, 50) = (³⁰C₁₀)(²⁰C₅)/(⁵⁰C₁₅)

= (30,045,015)(15,504)/(2,250,829,575,120)

P(X = 10) = 0.207

Hope this Helps!!!

7 0
2 years ago
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