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3 years ago

7

What is the quotient? 1462 divided 58 Enter your answer as a mixed number in the simplest form

2 answers:

stepan [7]3 years ago

8 0

The answer would be 25.2or 126/5

Vinil7 [7]3 years ago

3 0

Answer:

$25\frac{6}{29}$

Step-by-step explanation:

We are asked to find the quotient of 1462 divided 58 as a mixed number.

Write as a division problem:

$\frac{1462}{58}$

Reduce fraction:

$\frac{2\times 731}{2\times 29}$

$\frac{731}{29}$

$\frac{29\times 25+6}{29}$

$25\frac{6}{29}$

Therefore, our required quotient is $25\frac{6}{29}$ .

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1. Solve for the x and y intercepts:

aleksandrvk [35]

To find the x-intercept, you simply make y equal to 0 and solve the equation for x. To find the y-intercept, you make x equal to 0 and solve for y. So, for number 1, the answer is D because when you set it to y equals 0, you get x=3 and when you set it to x equals 0, you get 4.
Number 2 we would do the same thing and get D as well.
Hope this helps!

4 0

3 years ago

I need help please help me ;-;

vodomira [7]

Answer:

I think option A is correct answer

3 0

2 years ago

Read 2 more answers

Easy money : Explain how to muliply <br> 4 X 4 <br> 5 X 5<br> and <br> 6 X 6

defon

4 x 4 = 16
5 x 5 = 25
6 x 6 = 36

4 0

2 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

Which function has the smallest y-intercept value?

Lerok [7]

Answer:

Could you... put it normally?

Step-by-step explanation:

It's kinda confusing to read, so much understand

8 0

3 years ago