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3 years ago

579/3 579 divides by 3

Mathematics

2 answers:

dezoksy [38]3 years ago

6 0

The answer to the question is 193

Tpy6a [65]3 years ago

5 0

If you're asking what 579 divides by 3 is, then the answer is 193.

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In the diagram, the circle will be dilated by a scale factor of 3 about the origin. The points C, A, and B map to C', A', and B'

sp2606 [1]

Answer:

The correct answer is B. 15 units

Step-by-step explanation:

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2 years ago

Steve is buying apples for the fifth grade each bag holds $12 if there are 75 students total how many bags of apples will Steve

schepotkina [342]

Steve will have to buy "6.25" bags of apples if he wants to give one apple to each student.
Explanation:
75 divided by 12 = 6.25

8 0

3 years ago

Pls find the volume :))

vovangra [49]

Answer:

The volume can be found by multiplying these two numbers, which is

65/4 x 1/2

65/8 units^3

Let me know if this helps!

3 0

3 years ago

Read 2 more answers

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:

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

7 0

3 years ago

Find the area of a regular hexagon

xxMikexx [17]

<h2>Answer:</h2><h2>42 cm^2</h2>

please see the attached picture for full solution

hope it helps...

Good luck on your assignment....

4 0

3 years ago