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3 years ago

Select all statements that are true about the linear equation.

Mathematics

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

The correct Option are B and C.

Step-by-step explanation:

Consider the provided linear equation.

y=−2x+4

Consider the provided options.

Option A)

The graph of the equation is the set of values which satisfy the equation.

The graph of the equation is only one line, whereas the set of all points on the coordinate plane means the whole plane.

Thus, the option A) is False.

Option B)

The solution to the equation means the set of values which satisfy the equation.

Thus, the option B) The graph of the equation is the set of points that are solutions to the equation is correct.

Option C) The point (1, 2) lies on the the graph of the equation.

Substitute x=1 and y=2 in the provided equation.

2=−2(1)+4

2=−2+4

2=2

Which is true.

Hence, the option C) The point (1, 2) lies on the the graph of the equation is correct.

Option D) The point (0, 2) lies on the graph of the equation.

Substitute x=0 and y=2 in the provided equation.

2=−2(0)+4

2=4

Which is false.

Hence, the option D) The point (1, 2) lies on the the graph of the equation is incorrect.

Andre45 [30]3 years ago

4 0

The graph is all the solutions to the equation bthat statement is true

the point ,(1,2) does lie on the graph

the other 2 are false

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Step-by-step explanation:

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$18 = 14 + 0.20(t)

- 14

4 = 0.20t

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Answer:

The average hold time is 10.47 minutes.

Step-by-step explanation:

Let X = time the customers of a phone-in technical support business spend on hold.

The population mean of the random variable X is, μ = 11 minutes.

The population standard deviation of the random variable X is, σ = 1.16 minutes.

A random sample size, n = 62 callers are selected.

According to the Central limit theorem if large samples (n > 30) are selected from an unknown population with mean μ and standard deviation σ then the sampling distribution of sample mean ( $\bar x$ ) follows a Normal distribution.

The mean of the sampling distribution of sample mean is:

$\mu_{\bar x}=\mu=11$

The standard deviation of the sampling distribution of sample mean is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.16}{\sqrt{62}}=0.147$

It is provided that $P(\bar X>a)=0.79$ .

Compute the value of a as follows:

$P(\bar X>a)=0.79\\P(Z>z)=0.79\\1-P(Z$

The value of z for the above probability is, z = -0.806.

The value of a is:

$z=\frac{a-\mu_{\bar x}}{\sigma_{\bar x}}\\-0.806=\frac{a-11}{0.147}\\a=11-(0.86\times 0.147)\\a=10.87$

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