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blagie [28]
3 years ago
7

Please help me, I would gladly appreciate it. :)

Mathematics
1 answer:
Tems11 [23]3 years ago
7 0
A. Yes it is a function because all the x s are different

B If one of the x s repeated it would not be a function. Example: 4 8

C. Find slope. 7-5/4-3
Slope. 2/1. Or 2

Y=mx+b

7= 2(4)+b
7=8+b
B=-1

Y=2x-1
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Julli [10]
It helps to use your factor pairs
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3 years ago
( Math ) Please explain how you got this answer! Please use substitution method!
disa [49]

Answer:

y = 1

x = -2

Step-by-step explanation:

During the substitution method, you are substituting one function into another. Because both functions are set equal to "x", you can make the functions equal each other. This allows you to 'get rid' of the "x" variable and simplify to find the value of "y".

5A) x = -2y

5B) x = 2 - 4y

-2y = 2 - 4y                     <----- Set both equations equal

2y = 2                             <----- Add 4y to both sides

y = 1                                <----- Divide both sides by 2

Now that you know what "y" equals, you can plug that value into the "y" variable into one of the equations. This will allow you to then simplify and find the answer for "x".

x = -2y                             <----- Equation 5A

x = -2(1)                           <----- Plug 1 into "y"

x = -2                              <----- Multiply -2 and 1

5 0
2 years ago
What does the statement |b|=7<br> mean?
Umnica [9.8K]
The absolute value of b is equal to 7. Absolute value means the distance from zero. In simple terms, the positive version of the number. B could be 7 or -7, since both will be equal to 7 when you take their absolute value.

7 0
3 years ago
Read 2 more answers
Please help!!!! DIVIDING ALGEBRAIC EXPRESSIONS 2p/4p^2-1 divided by 6p^3/6p+3
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2p/(4p²-1)÷<span>6p</span>³<span>/(6p+3)
=</span>2p/(4p²-1) × (6p+3)/6p³
2p/(2p+1)(2p-1) ×3(2p+1)/6p³
=1/[(2p-1)*p²]=1/(2p³-p²)
5 0
4 years ago
A population of plastic chairs in a factory has a weight's mean of 1.5 kg and a standard deviation of 0.1 kg . Suppose a sample
Firlakuza [10]

Answer:

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 1.5, \sigma = 0.1, n = 100, s = \frac{0.1}{\sqrt{100}} = 0.01

What is the probability that the sample mean will be within +0.02 of the population mean?

Sample mean between 1.5 - 0.02 = 1.48 kg and 1.5 + 0.02 = 1.52 kg, which is the pvalue of Z when X = 1.52 subtracted by the pvalue of Z when X = 1.48. So

X = 1.52

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

By the Central Limit Theorem

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

Z = \frac{1.52 - 1.5}{0.01}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 1.48 ​

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

Z = \frac{1.48 - 1.5}{0.01}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +0.02 of the population mean.

3 0
3 years ago
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