Answer:
15 gallons
Step-by-step explanation:
if 6gallons = 192 miles
then x gallons 480 miles
x = (6*480)/192
x = 2880/192
x = 15 gallons
Answer:
8% of 800 is 64
800-64=736
Step-by-step explanation:
Answer:
length of 1 side of A, using the Pyth. Thm. and the dimensions of the other two squares: (side of A)^2 = (10 in)^2 + (24 in)^2. Then:
(side of A)^2 = 100+ 576 in^2 = 676 in^2.
Here I have not bothered to solve for the length of the side of A, since we want the area of square A. But if you do want the side length, find it: sqrt(676) = 26 in. Then the area of A is (26 in)^2 = 676 in^2.
Then the area of square A is (26 in)^2 =
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Step-by-step explanation:
Answer:
use PEDMAS
P: PARENTHESIS
E: EXPONENTS
D: DIVISON
M: MULTIPLICATION
A: ADDITION
S: SUBTRACTION
Step-by-step explanation:
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Answer:

Step-by-step explanation:
Given
Poisson Distribution;
Average rent in a week = 2.3
Required
Determine the probability of renting no more than 1 apartment
A Poisson distribution is given as;

Where y represents λ (average)
y = 2.3
<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>
<em />
Using probability notations;

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]




Solving for P(X = 1) [substitute 1 for x and 2.3 for y]









Hence, the required probability is 0.331