Answer:
a) what is the probability that Neither will of these products launch ?
= 0.30
b) At least one product will be launched ?
= 0.70
Step-by-step explanation:
From the above question, we have the following information:
P(A) = 0.45
P(B) = 0.60
P(A ∩ B) = P(A and B) launching = 0.35
Step 1
We find the Probability that A or B will launch
P (A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 0.60 + 0.45 - 0.35
= 1.05 - 0.35
= 0.70
a) what is the probability that Neither will of these products launch ?
1 - Probability ( A or B will launch)
= 1 - 0.70
= 0.30
b)At least one product will be launched?
This is equivalent to the probability that A or B will be launched
P (A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 0.60 + 0.45 - 0.35
= 1.05 - 0.35
= 0.70
Answer:
first box (x): 10
second box (y): 12
Step-by-step explanation:
HIJ is a translation of HIJ
translation just means you move the whole thing without any rotations dialations ect just moving it
from the blue shape it is asking what do you have to do to the x and y to get to the purple shape
you have to go +_x and take away _y
when you add a digit to x it moves to the right
when you add a digit to y it goes one up
because the HIJ are equaly spaced it dosnt matter which one we choose cause it will have the same answer
im going to use J
on the blue J you can see the x value is -8
the purple J's value is 5
to get from -5 to 5 you add 10
-5+5=0
0+5=5
5+5=10
therefore the first box (x) is 10
same thing for the y value im going to use J again
blue y value of J is 7
purple y value of J is -5
their distance is 12
7-7=0
0-5=-5
7+5=12
second box (y) is 12
i hope this helps
Answer:
5x^2(x^2 + 2x + 3)
Step-by-step explanation:
All the terms can be factored by 5 and x^2, so we get:
5x^2(x^2 + 2x + 3)
there are no more common terms so this is the answer
Answer:
Step-by-step explanation:
28 = 2 * 2 * 7
35 = 5 * 7
The common factor is 7 . This is the GCF
28 + 35 = 7(4 + 5) (answer)
Answer:
Estimates: $65
Actually: $73
65 / 73
= 0.9 (round to nearest tenth)
I hope this helps, I may be wrong. Somebody can correct me if I am wrong. :)
