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4 years ago

Determine whether each relationship is a function. State the domain and range of each relation.

Mathematics

1 answer:

atroni [7]4 years ago

3 0

<h2>Hello!</h2>

The answer is:

The relation is a function

Domain: {-5,-3,1}

Range: {6,3}

It's a function since each input (domain) has only one value at the output (range). A function exists when there is only a single value for each input, if there is more than one output for each input, the relation is not a function.

Have a nice day!

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Select all the expressions that are equivalent to 4-3.

Leviafan [203]

Answer:

See below.

Step-by-step explanation:

4*-3

A) -12

C) 2*-6

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3 years ago

The area of a square window is 0.25 square meter. How long is each side of the window in meters? *

mart [117]

Answer:

o.25

Step-by-step explanation:

Because a square has equal sides so all sides have to be 0.25

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3 years ago

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What is the constant of proportionality in the equation y=25X?

Ad libitum [116K]

Answer:

Step-by-step explanation:

y = kx

If y = 25x , then k = 25

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3 years ago

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wolverine [178]

Answer:

Step-by-step explanation:

150 -12.2 -rounded

350 -18.4.

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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

4 years ago