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podryga [215]
3 years ago
5

as you did during the lesson, label and represents the product or quotient by drawing disks on the place value chart 10 x 4 thio

usands equal

Mathematics
1 answer:
Naily [24]3 years ago
6 0

Answer : 40,000

Explanation : when multiplying 10 with 4,000 we get

4000 X 10 = 40,000.

As we multiply, using the chart there are 4 thousand present in the row of thousand column. When we multiply that with 10 it gets shifted to left and becomes 40,000.

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A board game that normally costs $30 is on sale for 25 percent off. What is the sale price of the game?
Paha777 [63]

We can use two different ways to solve this equation:

Because the board is 25 off, we can multiply the price and subtract the result:

30 - 0.25(30) = 30 - 7.5 = $22.50

We can also solve by multiplying the total price by 0.75

(1 - 0.25)(30) = 0.75(30) = $22.50

The sales price is $22.50

4 0
3 years ago
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Please help this is due after 30 minutes!
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Step-by-step explanation:

I entered it in and it was .9993319736

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I need help with this question. can someone please help me
Ahat [919]
-2x + y =3
use 5 number and put it in above equation to find y, I will use -2,-1,0,1,2
x      y
-2   -1
-1    1
 0    3
 1    5
 2    7

Then draw the graph, find another point in the graph which intersect the slope, 
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to prove it, i put (-3,-3) in the equation -2x + y = 3
and i got 
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7 0
3 years ago
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

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3 years ago
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1/6 since there are 6 sides, therefore, there is a 1/6 chance that any number will be rolled
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