16. 5x^3 y^-5 • 4xy^3
20x^4y^-2
20x^4 • 1/y^2
=20x^4/y^2
17. -2b^3c • 4b^2c^2
= -8b^5c^3
18. a^3n^7 / an^4 (a^3 minus a = a^2 same as n^7 minus n^4 = n^3)
=a^2n^3
19. -yz^5 / y^2z^3
= -z^2/y
20. -7x^5y^5z^4 / 21x^7y^5z^2 (divide -7 to 21 and minus xyz)
= -z / 3x^2
21. 9a^7b^5x^5 / 18a^5b^9c^3
=a^2c^2 / 2b^4
22. (n^5)^4
n ^5 x 4
=n^20
23. (z^3)^6
z ^3 x 6
=z^18
Answer:
sparky is 4 years older than cookie
__________________________
hope this helps, I wasn't entirely certain what the question was.
Answer:
5
Step-by-step explanation:
<u>move constant</u>
x+7 ≥ 12
<u>subtract</u>
x≥ 12-7
<u>solution</u>
x ≥ 5
The answer is 45 have a nice day
The probability that the sum of Michelle's rolls is 4 is 0.083
∴ P(A)=0.083
Step-by-step explanation:
Given that Michelle is rolling two six-sided dice, numbered one through six.
To find the probability that the sum of her rolls is 4:
∴ n(s)=36
Let P(A) be the probability that the sum of her rolls is 4
Then the possible rolls with sums of 4 can be written as
n(A)=3
The probability that the sum of her rolls is 4 is given by
=0.083
∴ P(A)=0.083
∴ the probability that the sum of Michelle's rolls is 4 is 0.083
-please give brainliest if correct!
