The question is incomplete. Here is the complete question:
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit 10 such targets in a row.
Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?
Answer:
40.13%
Step-by-step explanation:
Let 'A' be the event of not missing a target in 10 attempts.
Therefore, the complement of event 'A' is 
Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.
Now, 
We know that the sum of probability of an event and its complement is 1.
So, 
Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.
A is the answer 7th graders are not likely
Volume of the box= <span>56 cubic inches
let x is the length, then
width =</span><span>2 inches shorter than its length = x - 2
</span>height = <span>3 inches taller than its length = x+3
Volume = length x width x height
56 = x x (x-2) x (x+3)
56 = (x</span>² -2x)(x+3)
56 = x³ +3x² -2x² - 6x
56 = x³ + x² -6x
x³+x²-6x-56 = 0
using the rational root theorem and factoring the polynomial;
(x-4)(x² +5x +14) = 0
from here;
x-4 = 0
x = 4
So, length = 4 inches
width = x - 2 = 4 -2 = 2 inches
length = x + 3 = 4 + 3 = 7 inches
volume = l x w x h = 4 x 2 x 7 = 56
Answer:
its 100
Step-by-step explanation:
Step-by-step explanation:
3x+6+4x-20=10
7x-14=10
7x=24
x=24/7