Solution: We are given:
Let 
be the weight (oz) of laptop
We have to find 
To find the this probability, we need to find the z score value.
The z score is given below:
Now, we have to find 
Using the standard normal table, we have:
0.9236 or 92.36% of laptops are overweight
Answer:
-$13.5
Step-by-step explanation:
Let x be a random variable of a count of player gain.
- We are told that if the die shows 3, the player wins $45.
- there is a charge of $9 to play the game
If he wins, he gains; 45 - 9 = $36
If he looses, he has a net gain which is a loss = -$9
Thus, the x-values are; (36, -9)
Probability of getting a 3 which is a win is P(X) = 1/6 since there are 6 numbers on the dice and probability of getting any other number is P(X) = 5/6
Thus;
E(X) = Σ(x•P(X)) = (1/6)(36) + (5/6)(-9)
E(X) = (1/6)(36 - (5 × 9))
E(X) = (1/6)(36 - 45)
E(X) = -9/6 = -3/2
E(X) = -3/2
This represents -3/2 of $9 = -(3/2) × 9 = - 27/2 = -$13.5
Answer:
w= 6
Step-by-step explanation:
I just started by making an educated guess using the values already given. Then I inserted that into the problem to see if it worked.
L= 2w - 5
I used 6 as a random, educated guess for the value of w.
L = 2(6) - 5
L = 12-5
L = 7
Then, multiply L by 2 to account for both side lengths of the rectangle.
7(2)= 14
Subtract that value from the total perimeter to find what the width must equal.
26 - 14 = 12
Divide that answer by 2 since there are two sides for width.
12/2 = 6
I know this was kind of long, but I hope it helps! :)
