Answer:
11.25
Step-by-step explanation:
4.5 / 2 = 2.25 cups/ cup rice
2.25 x 2 cups of rice = 4.5 cups of water.
5 x 2.25 = 11.25
Simple. Just add/subtract each term. 4n-2n=2n. 8m+7m=15m.
This is very basic stuff.
Answer:
12
Step-by-step explanation:
Let's put the equations in standard form. For the first equation, we have:
−11y=6(z+1)-13y
2y−6z=6
y−3z=3
The second equation is:
4y−24=c(z−1)
4y−cz=24−c
If we multiply the first equation by 4, we get:
4y-12z=12
Comparing the two equations, we see that if c=12, both equations will be the same and there will be infinitely many solutions.
The correct value of c is 12.
Answer:
28x
Step-by-step explanation:
i took the test
Answer:
The probability that an athlete chosen is either a football player or a basketball player is 56%.
Step-by-step explanation:
Let the athletes which are Football player be 'A'
Let the athletes which are Basket ball player be 'B'
Given:
Football players (A) = 13%
Basketball players (B) = 52%
Both football and basket ball players = 9%
We need to find probability that an athlete chosen is either a football player or a basketball player.
Solution:
The probability that athlete is a football player = ![P(A)= \frac{13}{100}=0.13](https://tex.z-dn.net/?f=P%28A%29%3D%20%5Cfrac%7B13%7D%7B100%7D%3D0.13)
The probability that athlete is a basketball player = ![P(B)= \frac{52}{100}=0.52](https://tex.z-dn.net/?f=P%28B%29%3D%20%5Cfrac%7B52%7D%7B100%7D%3D0.52)
The probability that athlete is both basket ball player and football player = ![P(A\cap B) = \frac{9}{100}=0.09](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%20%3D%20%5Cfrac%7B9%7D%7B100%7D%3D0.09)
We have to find the probability that an athlete chosen is either a football player or a basketball player
.
Now we know that;
![P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3D%20P%28A%29%20%2B%20P%28B%29%20-%20P%28A%20%5Ccap%20B%29%5C%5C%5C%5CP%28A%5Ccup%20B%29%20%3D%200.13%2B0.52-0.09%3D0.56%5C%5C%5C%5CP%28A%5Ccup%20B%29%20%3D%20%5Cfrac%7B0.56%7D%7B100%7D%3D56%5C%25)
Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.