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katrin2010 [14]

3 years ago

11

Six girls were playing a game with marbles. The 48 marbles were to be divided evenly among the girls. How many marbles did each

girl receive?

1 answer:

horrorfan [7]3 years ago

8 0

Each girl received 8 marbles.

48÷6 = 8

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An ice cream store sells 3drinks, in 4sizes, and 8 flavors. In how many ways can a customer order a drink?

Sedbober [7]

You would take 3 x 4 which equals 12. And then you would take 12 x8 which would equal 96

7 0

4 years ago

line y = 2x+1 is transformed by a dialation with a scale factor of two and centered at (0,-4). what is the equation of the image

Gekata [30.6K]

Answer: y = 2x + 6

Step-by-step explanation:

Here the equation of line,

y = 2x + 1

y-intercept of the line is, (0, 1)

And, the slope of the line is 2.

Since, If a point (x,y) is dilated by a scale factor k about a point (a,b),

Then transformed point are get by,

$(x,y)\rightarrow (k(x-a)+a, k(y-b)+b)$

Thus, The transformed point of point (0,1) when dilation is occur by the scale factor 2 about the point (0,-4),

$(0,-4)\rightarrow (2(0-0)+0, 2(1+4)-4)$

$(0,-4)\rightarrow (0, 6)$

Thus, the point of the new line is (0,6)

And, the slope of the new line is 2 ( because in the dilation, the slope of the line does not change)

Thus, the equation of new line,

(y - 6) = 2 (x - 0)

y - 6 = 2x

y = 2x + 6

5 0

3 years ago

Solve the equation!! <br> 6 + log9 -5n = 6

scoray [572]

Answer:

n=log(9 1/5)

Step-by-step explanation:

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3 years ago

Does each function describe exponential growth or decay?

zhannawk [14.2K]

Hey Jinx :)

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Note that a decay has a rate between 0 and 1, and a growth as a rate of 1 or higher.

y = 1/2(1 + 0.03)^t → y = 1/2(1.03)^t (Rate is more then 0) GROWTH

y = 0.3(0.95)^t → (Rate is between 0 and 1) DECAY

y = ((1+0.03)^1/2)^2t → y = y = 1.014^2t (Rate is more then 0) GROWTH

y = 200(0.73)^t → (Rate is between 0 and 1) DECAY

y = 4(1/4)^t → (Rate is between 0 and 1) DECAY

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Hope This Helped! Good Luck!

4 0

3 years ago

Three six-sided fair dice are rolled. The six sides of each die are numbered 1; 2; : : : ; 6. Let A be the event that the first

valkas [14]

Answer:

1.) [A, B, C]

2.) [A', B', C']

3.) [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

4.) A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

Step-by-step explanation:

If A is the event that the first die shows an even number, then A' is the event that first die DOES NOT show an even number.

If B is the event that the Second die shows and even number, then B' is the event that second die DOES NOT show an even number.

If C is the event that the third die shows an even number, then C' is the event that third die DOES NOT show an even number.

Hence, our possible events are A, A', B, B', C, C'.

For question 1, in the event that all three dice show and even number, the expression of the event = [A, B, C]

For question two, in the event that no die shows an even number, the expression of the event = [A', B', C']

For question 3,In the event that at least one die shows an odd number, the expression of the event = [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

For question 4,in the event that at most two dice show odd numbers, the expression of the event = [A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

6 0

3 years ago