What is the prime factorization of 195
2 answers:
Answer: 5 x 3 x 13
Step-by-step explanation: To find the prime factorization of 195, first create a factor tree with two branches.
Since 195 end in 5, we begin by dividing 195 by 5 which gives us 39 so this means that 5 and 39 are factors of 195. This means that we write a 5 and 39 at the bottom of the branches. Since 5 is a prime number, we circle the 5.
Since 39 is not prime, we draw two new branches. 39 divided by 3 is 13 so 3 and 13 are factors of 39 which means that we write 3 and 13 at the bottom of the branches. Since both 3 and 13 are prime, we circle both numbers.
Now, since we have circles at the bottom of all of or branches, we are finished.
So the prime factorization of 195 is 5 × 3 × 13.
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Answer:
y+3=2(x-1)
Step-by-step explanation:
Point-slope form is . In order to write a linear equation using this form, the
,
and
must be substituted by real numbers.
The represents the slope of the equation. We know that the answer has to be parallel to the line y=2x-1. y=2x-1 is already in slope-intercept form, or
, in which
represents the slope. We can see that the
is in place of that
, therefore 2 is the slope of y=2x-1. Lines that are parallel have the same slope, so we know that the slope of the new equation must be 2 as well.
The and
represent the x and y values of one point that the line passes through. We know that the new equation must pass through the point (1, -3), so we'll substitute 1 in place of the
and -3 in place of the
.
Substituting for those three values will give you an equation like this, therefore it is the answer:
Answer:
(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.
(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.
(c) The probability that exactly 1 of the next three vehicles passes is 0.189.
(d) The probability that at most 1 of the next three vehicles passes is 0.216.
(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.
Step-by-step explanation:
Let <em>X</em> = number of vehicles that pass the inspection.
The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.
(a)
Compute the probability that all the next three vehicles inspected pass the inspection as follows:
P (All 3 vehicles pass) = [P (X)]³
Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.
(b)
Compute the probability that at least 1 of the next three vehicles inspected fail as follows:
P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)
Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.
(c)
Compute the probability that exactly 1 of the next three vehicles passes as follows:
P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)
= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)
+ P (Only 3rd vehicle passes)
Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.
(d)
Compute the probability that at most 1 of the next three vehicles passes as follows:
P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)
+ P (0 vehicles passes)
Thus, the probability that at most 1 of the next three vehicles passes is 0.216.
(e)
Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.
Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:
Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.