Area=basexheight
Area=9x4
Area=36
Answer:
V =41.41³
A = 94.41²
----
V =225.16³
SA =283.25²
----
V = 64³
SA =113.32²
----
V =433.33³
SA = 378.57²
Step-by-step explanation:
Picture 2 = a = 1/2 base = 3.5 x 3.5 = 12.25 b= 5 x 5 = 25
c²= a² + b² = 3.5² + 5²
c ²= √12.25 + √25
c ²= √ 37.5 = 6.12372435696
c ² = 6.1237 missing side
Picture 1 + 2 formula SA = bh + (s1 + s2 + s3)H
V = V= 1/2 b x h h x SA
Picture 3 + 4 formula SA= a²+ 2a a² / 4 + h² V= a² h/3
Answer:
8.6 × 10-9
Step-by-step explanation:
<h2>
Answer:</h2>
-18x^5-204x^4-722x^3-600x^2+672x-128
<h2>
Explanation:</h2>


There are multiple ways to solve this for example the foil method or the box method. I prefer to do the box method.
First we are going to set it up
After we set it up we have to solve it so first we are going to multiply

We have to add the exponent




Now we have to add them
-18x^5 stays because there is nothing to add to
Now we add the rest



And -128 stays the same

The answer is
-18x^5-204x^4-722x^3-600x^2+672x-128
Answer:
The proportion of student heights that are between 94.5 and 115.5 is 86.64%
Step-by-step explanation:
We have a mean
and a standard deviation
. For a value x we compute the z-score as
, so, for x = 94.5 the z-score is (94.5-105)/7 = -1.5, and for x = 115.5 the z-score is (115.5-105)/7 = 1.5. We are looking for P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664. Therefore, the proportion of student heights that are between 94.5 and 115.5 is 86.64%