Answer:
a. 2/5
b. 1/2
c. 100/9
Step-by-step explanation:
a: Multiply bottom and top by 100 to keep values the same and make the fraction a whole fraction:
.12 × 100 = 12
.3 × 100 = 30
Simplify: 12/30 = 2/5
b: Multiply bottom and top by 1,000 to keep values the same and make the fraction a whole fraction:
.125 × 1,000 = 125
.25 × 1,000 = 250
Simplify: 125/250 = 1/2
c: Multiply bottom and top by 100 to keep values the same and make the fraction a whole fraction:
4 × 100 = 400
.36 × 100 = 36
Simplify: 400/36 = 100/9
Answer:
The Lateral Surface Area is 911.32 square unit and
Total Surface Area is 1361.7 square unit.
Step-by-step explanation:
For a given Cone
Radius (r) = 12
Height (h) = 21
<u>For Lateral Surface Area</u>
<h3>
<u>Formula</u><u>:</u> </h3>
A = πr√r² + h²
A = 3.14 × 12 × √(12)² + (21)²
A = 3.14 × 12 × √144 + 441
A = 3.14 × 12 × √585
A = 37.68 × 24.18
A = 911.32 square unit
Now,
<u>For</u><u> </u><u>Total</u><u> Surface Area</u>
<h3><u>Formula:</u></h3>
A = πrl + πr²
For Slant height (l)
l² = r² + h²
l² = (12)² + (21)²
l² = 144 + 441
l² = 585
l = 24.18
So,
A = πrl + πr²
A = πr(l + r)
A = 3.14 × 12 × (24.14 + 12)
A = 3.14 × 12 × 36.14
A = 1361.7 square unit
Thus, The <u>Lateral Surface Area</u> is 911.32 square unit and <u>T</u><u>otal Surface Area</u> is 1361.7 square unit.
<u>-TheUnknownScientist</u>
Answer:
Step-by-step explanation: Given : f(x)=x
4
−3x
2
+4
f(2)=(2)
4
−3(2)
2
+4
=16−12+4=8
The answer is 220 because you would just have to round 1 to 2 than everytging behind it is a zero
Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
