The way we simplify this is by using the distributive property and multiply every term by 0.5. When we do this we get 2a+3b
(<u>−1</u>
2 )(n^3)+
<u>1</u>
2 n^2+4.6n+(−
<u>1</u>
2)(n^3)+
<u>1</u>
2 n^2+4.5n
=
<u>−1</u>
2 n^3+
1
2 n^2+4.6n+
−1
2 n^3+
1
2 n^2+4.5n
Combine Like Terms:
=
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.6n+
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.5n
=(<u>−1</u>
2 n^3+
<u>−1</u>
2 n^3)+(
<u>1</u>
2 n^2+
<u>1</u>
2 n^2)+(4.6n+4.5n)
=−n^3+n^2+9.1n
Answer:
=−n^3+n^2+9.1n
Everything underlined means its a fraction/divided hope this helps <em>:D</em>
Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
Answer:
point a
Step-by-step explanation:
(0,-7) would be left 7 units so its point a