Question:
A cafeteria offers oranges, apples, or bananas as its fruit option. It offers peas, green beans, or carrots as the vegetable option. Find the number of fruit and vegetable options. If the fruit and the vegetable are chosen at random. what is the probability of getting an orange and carrots? Is it likely or unlikely that a customer would get an orange and carrots?
Answer:
The probability of getting an orange and carrots is 
Step-by-step explanation:
The fruits offered in cafeteria = oranges, apples, or bananas.
The vegetables offered in cafeteria = peas, green beans, or carrots.
There are 3 fruits and 3 vegetables . Therefore the total possible number of outcomes is =
= 9
Now the probability of getting an orange and carrots = 
The possibility is very unlikely to happen
Combine like terms, 1+6 is 7 and 2t+4t is 6t. so 6t+7=55. get the t by itself by subtracting 7 from both sides. 6t=48, divide by 6. t=8
Answer:
0
Step-by-step explanation:
To solve this, you first plug in the 4 and 2 where the x and y's go, making the expression (4 - 2(2))*(4 + 2(2)). Then you solve each parenthesis (below).
(4 - 2(2))*(4 + 2(2)).
(4 - 4)*(4 + 4).
0*8
0
Hope this helped! If you have any questions, feel free to message me :)
If you would like to find the x-intercepts of the function f(x) = - 2 * x^2 - 3 * x + 20, you can calculate this using the following steps:
f(x) = - <span>2 * x^2 - 3 * x + 20
</span>f(x) = - (2x - 5) * (x + 4)
1. x = - 4
2. x = 5/2
(x, y) = (-4, 0)
The correct result would be (-4, 0).
7 multiplicado por 49 =343