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3 years ago

This question is about Domain and Range. plz don't answer if u don't know.

Mathematics

1 answer:

drek231 [11]3 years ago

5 0

no.1 is -4

no.2 is -2

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the perimeter of a square is equal to the perimeter of an equilateral triangle the length of a side of the square is given by th

diamong [38]

Answer:

x = 3

Step-by-step explanation:

We are told that the length of a side of the square is x.

Now, perimeter of square = 4 × side length

Perimeter of square = 4x

Also,we are told that the triangle is equilateral and a side is (x + 1).

Thus, perimeter of triangle = 3(x + 1)

Since we are told that the perimeter of the square is equal to that of the equilateral triangle. Thus;

4x = 3(x + 1)

4x = 3x + 3

Subtract 3x from both sides to get;

4x - 3x = 3

x = 3

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3 years ago

What are some limitations in using a line of best fit drawn on a scatter plot to make predictions?

denis23 [38]

Answer: It is possible to draw different lines to approximate the same data. The line of best fit is only an estimate.

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2 years ago

A - (a)(a - c); use a = 6, and c = 6<br> Evaluate using the values given

pishuonlain [190]

Answer:

6-(6)(6-6) = 6

Step-by-step explanation:

6-6 is 0

6*0 is 0

6-0 is 6

6 is answer

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3 years ago

Does anyone know how to 'correctly' input the answers into mathcentre?

klio [65]

Put the individuals in the brackets

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3 years ago

Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue. We randomly select 5 marbles from

Aleonysh [2.5K]

Answer:

probability of selecting 3 green marbles, 1 red marble, and 1 blue marbles

$P(E) =\frac{8}{126} = 0.0634$

Step-by-step explanation:

Given data urn containing '9' marbles

given red marbles = 2

green marbles =3

blue marbles = 4

Five marbles can be selected at a time from '9' marbles in $9_{C_{5} }$ ways

by using formula $n_{C_{r} }=\frac{n!}{(n-r)!r!}$

$9_{C_{5} }=\frac{9!}{(9-5)!5!} = \frac{9X8X7X6X5!}{4!5!}$

After simplification , we get $9_{C_{5} } = 126$

total number of ways n(S) = 126

The probability of selecting '3' green marbles ,one red marble and one blue marble with replacement.

let ' E' be the event of selecting '3' green marbles ,one red marble and one blue marble with replacement.

n(E) = $3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }$ ways

The required probability $P(E) = \frac{n(E)}{n(S)}$

$p(E) = \frac{3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }}{9_{C_{5} } }$

on simplification , we get

$P(E) = \frac{1X2X4}{126} =\frac{8}{126}$

$P(E) =\frac{8}{126} = 0.0634$

7 0

4 years ago