Answer:
4 1/4
Step-by-step explanation:
2 1/2 + 1 1/4 + 1/2 = 5/2 + 5/4 + 1/2
LCM = 4
5/2 x 2/2 = 10/4
5/4 = 5/4
1/2 x 2/2 = 2/4
10/4 + 5/4 + 2/4 = 17/4
= 4 1/4
Therefore he put 4 1/4 cups altogether.
Answer:
≈ 542,029 copies
Step-by-step explanation:
37,400 copies represents a 6.9% of the total copies sold to date.
Therefore; Copies sold to date will be 100%
Thus;
Total number of copies sold = (37,400/6.9)× 100
= 542,028.99 copies
<u>≈ 542029 copies</u>
Here, You need to apply "Proportionality Theorem"
According to this theorem, the ratio of components of a leg would be equal to the ratio of corresponding components of another leg.
In mathematical form,
TV/VS = TW/WU
Substitute their values,
14/6 = 21/(x+4)
14(x+4) = 21*6
14x+56 = 126
14x = 126-56
x = 70/14
x = 5
In short, Your Answer would be 5
Hope this helps!
Given:
The number of male professors = 15.
The number of female professors = 9.
The number of male teaching assistants = 6.
The number of female teaching assistants = 12.
A person is selected randomly from the group.
Required:
We need to find the probability that the selected person is a professor or a male.
Explanation:
The total number of people in the group = 15+9+6+12 = 42
n(S) =The total number of people in the group

Let A be the event that the selected person is a professor or a male.
The number of people who are professors or male = 15+9+6 = 30
n(A)= The number of people who are professors or male.

Let P(A) be the probability that the selected person is a professor or a male.



Final answer:
The probability that the selected person is a professor or a male is 5/7.
Area inside the semi-circle and outside the triangle is (91.125π - 120) in²
Solution:
Base of the triangle = 10 in
Height of the triangle = 24 in
Area of the triangle = 

Area of the triangle = 120 in²
Using Pythagoras theorem,




Taking square root on both sides, we get
Hypotenuse = 23 inch = diameter
Radius = 23 ÷ 2 = 11.5 in
Area of the semi-circle = 

Area of the semi-circle = 91.125π in²
Area of the shaded portion = (91.125π - 120) in²
Area inside the semi-circle and outside the triangle is (91.125π - 120) in².