In y = mx + b form, the slope can be found in the m position and the y intercept can be found in the b position.
y = mx + b
y = 2x + 15
slope(m) = 2 and y int (b) = 15
The initial amount is the y intercept, which is 15. The rate of change (the slope) is 2
example :
baskets picked in 1 hr :
y = 2(1) + 15
y = 2 + 15
y = 17
baskets picked in 2 hrs :
y = 2(2) + 15
y = 4 + 15
y = 19
so the slope ( rate of change) is basically saying for every hr of picking, you pick 2 baskets...so your picking 2 baskets per hr.
The y intercept (15) is telling us that they already had 15 baskets to start with.
The answer is 2.36 :) :) :) :) :)
Let us solve this system of equations by using the elimination method. Adding the 2 equations, we get
9x + 5y - 9x + 4y = -33 + 6
9y = -27
y = -3
Substituting this value of y in the first equation, we get
-9x + 4(-3) = 6
-9x - 12 = 6
9x + 12 = -6
9x = -18
x = -2
Therefore, x = -2, and y = -3. Hope this helps! If you have any questions, feel free to ask.
Answer:
The total number of apples is 17
Step-by-step explanation:
(a)
Number of green apples = 4
P ( that all three apples are green)
= (limit 4 and 3) / (limit 17 and 3)
= 4! / 3!(4−3)!
= 17! / 3!(17−3)!
= 1 / 170
= 0.005
(b)
Number of red apples = 8
P(that all three apple are green)
=(limit 8 and 3) / (limit 17 and 3)
= 8! / 3!(8−3)!
= 17! / 3!(17−3)!
= 7 / 85
= 0.082
Now,
P(that no three apple is red)= 1−0.082 = 0.918
c)
P(of selecting 4 apple that contain at least 2 red apples)
= (limit 8 and 2) /( limit 17 and 2) X (2/15) + (limit 8 and 3) /(limit 17 / 3) X (1/14)
= (7/34) × (2/15) + (7/85) × (1/14)
= 0.027 + 0.058
= 0.085
d)
P(that the second apple selected is yellow given that the first apple is red)
= (limit 5 and 1) / ( limit 16 and 1)
= 5! / 1!(5−1)!
= 16! / 1!(16−1)!
= 5 /16
= 0.3125
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
