<h3>If this helped, PLEASE make this the BRAINLIEST answer!</h3>
0.4641. would be your answer
Adding up there are 41769 five digit numbers with atleast one zero in it. Therefore the probability that a five-digit number has at least one zero in it is 0.4641.
Factor the expression x-x^2y
Answer:
Step-by-step explanation:
I=PRT/100
Multiply both sides by 100
100I=PRT
divide both sides by the coefficient of P which is RT
PRT/RT= 100I/RT
P= 100I/RT
We are asked in this problem to simplify the statements written on the problem. In this case, we can start converting words into numbers into the function.
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1) (3/x + 2/(x+2) ) /(3/(x+2) -2/x. We should note that the word quantity denotes a parenthesis is started and that a set of terms is grouped inside the parenthesis.
simplifying, we try to make a common denominator out of this terms.
numerator:
denominator is x*(x+2), so the make up of the numerator is 3*(x+2) + 2x / (x*(x+2).
</span>denominator: 3x - 2(x-2) / (x*(x+2). <span>
The denominator has the same set of denominators which may cancel out both denominators. Hence,we simplify
num/denom = 3x+6 +2x / 3x -4x +4 = 5x +6 / 4-x
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120.0 hope his was helpful