Answer:
Step-by-step explanation:
We need to find the probability that the mechanic will service or more cars.
It's a simpler one given that we have the probabilities of servicing 4 or less cars.
P(at least 5 cars) is given by subtracting the probabilities of servicing both 3 and 4 cars.
Answer:
<em>24 square yards</em>
Step-by-step explanation:
Find the diagram attached.
The area of the diagram = Area of rectangle + Area of triangle
Area of rectangle = 3 * 4
Area of rectangle = 12 square yards
Area of triangle = 1/2 * base * height
Area of triangle = 1/2 * 6 * 4
Area of triangle = 12 square yards
Area of the diagram = 12 + 12
Area of diagram = 24 square yards
<em>Hence 24 square yards of carpeting is needed</em>
Let p = number of pennies.
Let n = number of nickels.
We are given that n= 2p and the total value is $8.80.
We know that a penny = $0.01 and that a nickel = $0.05.
So $0.01p + $0.05n = $8.80.
Substitute 2p for n:
$0.01p + $0.05*2p = $8.80
$0.01p + $0.10p = $8.80
$0.11p = $8.80
p = 80
So n = 2p = 2*80 = 160
Thus there are 80 pennies ($0.8) and 160 nickels ($8.00). The value of all the coins is $8.80.
Answer:
A. Pages about Michael Jordan.
Step-by-step explanation:
Michael Jordan, MJ, is a great basketball player, and has achieved one the best records in his career. He is a National Basketball Association (NBA) player with great skills and energy.
From the search item, the two names Michael OR Jordan is for one person, Michael Jordan. The search would combine the two names because it is a well known one and give an output on Michael Jordan. Thus, the pages that would be favored are pages about Michael Jordan.
Answer:
B
Step-by-step explanation:
Mathematically, we shall be using the z-score here
Z-score = (x-mean)/SD
From the question, mean = 80 and SD = 4
So we want to get the option in the question that has a z-score of 1 or below
Let’s look at 75
80-75 = 5 ( this is clearly above 1 standard deviation of 4)
For 77;
80 -77 = 3
This is less than the standard deviation of 4, meaning it is within 1 standard deviation of the mean
