All categories

3 years ago

PLS HELP QUICK I NEED THIS TO PASS

Mathematics

1 answer:

garri49 [273]3 years ago

6 0

Answer:

Add all numbers and you get 2160

Then 2160÷2= 1080

This is average price.

If I'm right so,

Please mark me as brainliest. thanks!!!

You might be interested in

What is .744 as a fraction

jenyasd209 [6]

744/1000 would be your answer

4 0

3 years ago

Read 2 more answers

Instructions: Solve the following equation for the variable given.

frutty [35]

Answer:

x = -1

Step-by-step explanation:

1/4 (4x - 4) = -2(6x + 7)

1x - 1 = -12x - 14

13x = -13

x = -1

7 0

3 years ago

A bus travels two different routes: the Green Route and the Blue Route.

Alex17521 [72]

Answer:4.5

Step-by-step explanation: Check the file for my explanation

8 0

2 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:

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

7 0

3 years ago

What is the interquartile range?<br> The interquartile range is __

xenn [34]

Answer:

<em>Your IQR or interquartile range is 2.</em>

Step-by-step explanation:

Median: 3

lower quartile: 2

upper quartile: 4

Interquartile range: upper quartile - lower quartile = answer

Interquartile range: 4 - 2 = 2

~hope this helps~

I would highly appreciate it if you give me the brainliest~ :)

4 0

3 years ago