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4 years ago

Trent began his math work at 5:29 p.m. He finished at 6:04 p.m. How many minutes did Trent spend on his math homework?

Mathematics

2 answers:

Morgarella [4.7K]4 years ago

7 0

Answer: I am pretty sure it is 35 min

Step-by-step explanation:

Inga [223]4 years ago

3 0

Answer

Trent spent 35 minutes on his work.

Step-by-step explanation:

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Answer:

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

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X 5 6 7 8 9 10

P(X) 0.05 0.15 0.33 0.28 0.12 0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

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

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :

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

And using the formula we got:

$E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229$

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