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1 year ago

Jason Reads 126 pages in 6 hours. Joe reads 115 pages in 5 hours how many more pages per hour does Joe read?

1 answer:

4 0

Jason --> 126/6 = 21 pages per hour

joe --> 115/5 = 23 pages per hour

joe reads 2 more pages per hour

hope this helps :)

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Can someone help please

riadik2000 [5.3K]

Answer:

62 Sunday papers were sold

Step-by-step explanation:

Let x represent the number of Sunday papers sold. Then half that many, or x/2, is the number of Friday papers sold. The total revenue from the sales is the sum of products of quantity and price:

1.50 · x + 0.75 · (x/2) = 116.25

Multiplying by 2, this becomes ...

3.00x +0.75x = 232.50

3.75x = 232.50

Dividing by the coefficient of x gives ...

232.50/3.75 = x = 62

The number of Sunday newspapers sold is 62.

6 0

2 years ago

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

8 0

2 years ago

An onion farmer is hiring workers to help harvest the onions. He knows that the number of days it will take to harvest the onion

Alisiya [41]

A relationship between a set of inputs

6 0

1 year ago

Read 2 more answers

6.2 ÷ 3.1<br> is equivalent to <br> 62 ÷ 31<br> The dividend and divisor were both multiplied by

VARVARA [1.3K]

Answer:

The dividend and the divisor were both multiplied by 10

Step-by-step explanation:

8 0

1 year ago

An elephant has a mass of 6.8 × 10 3 kg.

telo118 [61]

A). because B,C, and D didnt really make any sence

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1 year ago