1 apple + 1 banana
1 apple + 1 orange
1 apple + 1 peach
1 banana + 1 apple
1 banana + 1 orange
1 banana + 1 peach
1 orange + 1 apple
1 orange + 1 banana
1 orange + 1 peach
1 peach + 1 apple
1 peach + 1 orange
11 combinations of food
Hope this helped
<span>Exactly 33/532, or about 6.2%
This is a conditional probability, So what we're looking for is the probability of 2 gumballs being selected both being red. So let's pick the first gumball.
There is a total of 50+150+100+100 = 400 gumballs in the machine. Of them, 100 of the gumballs are red. So there's a 100/400 = 1/4 probability of the 1st gumball selected being red.
Now there's only 399 gumballs in the machine and the probability of selecting another red one is 99/399 = 33/133.
So the combined probability of both of the 1st 2 gumballs being red is
1/4 * 33/133 = 33/532, or about 0.062030075 = 6.2%</span>
Answer
(-3 * 2p) - (-3 * 3q)
Step-by-step explanation:
I dunno how to explain it but that's it
Step-by-step explanation:
f(x) = 4x + 1
g(x) = x² - 5
(f+g)(x) = f(x) + g(x)
= (4x + 1) + (x² - 5)
= x² + 4x + (1 - 5)
= x² + 4x - 4
Option → C