Write the following phrase as an algebraic expression. Use n for "a number."
1 answer:
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First of all, you add 1/3 and 4/5, so you get:
To add 2 fractions, they need to have the same denominator. Since 3 and 5 don't have a common factor, so you multiply 1/3 by 5, and 4/5 by 3, so you get:
Now, the two fractions got the same denominator, and you can add them by adding the numerators over the same denominator. (You don't add denominators in addition) so you get:
That's the sum of 1/3 + 4/5.
Now, they want the sum of the 2 fractions 7 times. So you multiply 17/15 by 7. And as you know, 7 = 7/1. So you get:
In multiplication no need to have the same denominator. You multiply nominators by nominators and denominators by denominators. So you get:
Since 119 and 15 don't have a common factor, so they can't be simplified.
So, the answer is 119/15.
Hope this Helps! :D
To add 2 fractions, they need to have the same denominator. Since 3 and 5 don't have a common factor, so you multiply 1/3 by 5, and 4/5 by 3, so you get:
Now, the two fractions got the same denominator, and you can add them by adding the numerators over the same denominator. (You don't add denominators in addition) so you get:
That's the sum of 1/3 + 4/5.
Now, they want the sum of the 2 fractions 7 times. So you multiply 17/15 by 7. And as you know, 7 = 7/1. So you get:
In multiplication no need to have the same denominator. You multiply nominators by nominators and denominators by denominators. So you get:
Since 119 and 15 don't have a common factor, so they can't be simplified.
So, the answer is 119/15.
Hope this Helps! :D
Answer:
a) 0.778
b) 0.9222
c) 0.6826
d) 0.3174
e) 2 drivers
Step-by-step explanation:
Given:
Sample size, n = 5
P = 40% = 0.4
a) Probability that none of the drivers shows evidence of intoxication.
b) Probability that at least one of the drivers shows evidence of intoxication would be:
P(X ≥ 1) = 1 - P(X < 1)
c) The probability that at most two of the drivers show evidence of intoxication.
P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)
d) Probability that more than two of the drivers show evidence of intoxication.
P(x>2) = 1 - P(X ≤ 2)
e) Expected number of intoxicated drivers.
To find this, use:
Sample size multiplied by sample proportion
n * p
= 5 * 0.40
= 2
Expected number of intoxicated drivers would be 2