Hey there! :)
Answer:
SA = 144 cm².
Step-by-step explanation:
Find the surface area by calculating the areas of each of the lateral sides and bases:
In this instance, the bases are triangles, so the formula A = 1/2(bh) will be used:
Bases:
A = 1/2(bh)
A = 1/2(4·3)
A = 1/2(12)
A = 6 cm².
There are two bases, so:
6 × 2 = 12 cm²
Find the areas of the lateral sides using A = l × w:
5 × 11 = 55 cm²
4 × 11 = 44 cm²
3 × 11 = 33 cm²
Add up all of the areas:
12 + 55 + 44 + 33 = 144 cm².
This kinda doesn’t make sense but maybe 80, 80, 20
Answer:
72.51
Step-by-step explanation:
Given that :
Lab score % = 23%
Each Major test % = 22.5%
Final exam % = 32%
Score :
Lab score = 96
First Major test = 64
Second major test = 62
Final exam = 69
Hence,
Weighted average = Σ(weight % * score)
Weighted average = Σweight_1*score_1 +.. Weight_n * score_n)
(0.23 * 96) + (0.225 * 64) + (0.225 * 62) + (0.32 * 69) = 72.51
Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
Answer:
d
Step-by-step explanation:
