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dexar [7]
3 years ago
15

Solve each system of equations by substitution. List the ordered pair.

Mathematics
1 answer:
Anit [1.1K]3 years ago
8 0
Y=6x-11
-2x-3y=-7
-2x-3(6x-11)=-7
-2x-18x+66=-7
-20x=-73
x=3.65
y=6(3.65)-11=10.9

y=-3x+5
5x-4y=-3
5x-4(-3x+5)=-3
5x+12x=17
x=1
y=-3(1)+5=2

2x-3y=-1
y=x-1
2x-3(x-1)=-1
-x=-4
x=4
y=4-1=3

-3x-3y=3
y=-5x-17
-3x-3(-5x-17)=3
12x=-48
x=-4
y=-5(-4)-17=3
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Step-by-step explanation:

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Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o
AysviL [449]

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean 490 mg and variance of 400.

This means that \mu = 490, \sigma = \sqrt{400} = 20

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{500 - 490}{20}

Z = 0.5

Z = 0.5 has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

4 0
3 years ago
Let x be normal distribution with mean 70 and standard deviation 2. given the following probability, p (x < a) = 0.9147, find
lapo4ka [179]

The value of a is 72.74.

<h3>What is Probability ?</h3>

Probability is the study of likeliness of an event to happen.

It is given that

The mean \rm \mu of the distribution is 70

Standard deviation \rm \sigma  is 2

p( x<a) = 0.9147

a = ?

From Z table

For p ( p < 0.9147) , z = 1.37

\rm Z = \dfrac { X - \mu}{ \sigma}

\rm \rm 1.37 = \dfrac { a - 70}{ 2}

a = 70+ 2*1.37

a = 72.74

Therefore the value of a is 72.74.

To know more about Probability

brainly.com/question/11234923

#SPJ1

6 0
2 years ago
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