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pav-90 [236]
3 years ago
8

A range value of the function f(x)=−2x+6 is 0. What is its domain value?

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
8 0

Answer:

The domain is all real numbers

-inf<x<inf

Thepotemich [5.8K]3 years ago
4 0

Answer:

3

Step-by-step explanation:

First, we need to understand what the problem is asking for.

The range of a function is the set of the y-coordinates of all the points of the function.

The domain of a function is the set of the x-coordinates of all the points of the function.

"A range value" means the same as "the y-coordinate of one point."

This problem is telling you that the y-coordinate of a point of function f(x) =  -2x + 6 is 0. Then it asks what its domain value is. It is asking you what is the x-coordinate of the point that has 0 as its y-coordinate.

Simply put, for function f(x) = -2x + 6, when y = 0, what is x?

f(x) = −2x + 6

-2x + 6 = -

-2x = -6

x = 3

Answer: 3

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Step-by-step explanation:

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8 0
3 years ago
A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard
Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

\sigma=\frac{0.8}{\sqrt{37}}=0.13

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, P(z > 1.69) = 1 - P(z < 1.69)

P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0
3 years ago
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