All categories

3 years ago

In an EFE Matrix, how much more important is a weight of 0.20 than a weight of 0.15? A. 50% B. 33% C. 25% D. .05% E. 5.0%

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer: B. 33%

Step-by-step explanation:

In Mathematics , a percentage is basically a process of dividing the numerator by the denominator , then multiply the answer by 100.

We use Mathematical Formula : $\dfrac{\text{Difference between the Weight of 0.20 and Weight of 0.15 }}{\text{Weight of 0.15}}\times100$

Since , Difference between the Weight of 0.20 and Weight of 0.15 = 0.20-0.15=0.05

Then, substitute the value in formula , we get

$=\dfrac{0.05}{0.15}\times100\\\\=\dfrac{5}{15}\times100\\\\ \dfrac{100}{3}\%\approx33\%$

Hence, the weight of 0.20 is 33% more important than a weight of 0.15.

You might be interested in

What is the value of a in this function's equation?<br> A. -1<br> B. 1<br> C. 2<br> D. -2

Pavlova-9 [17]

B is the correct answer !!!!

7 0

3 years ago

Answeeeerr please ????

valentinak56 [21]

It is the first option.

4 0

3 years ago

Read 2 more answers

Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

8 0

3 years ago

Which expression is equivalent to 7a^2b + 10a^2b^2 + 14a^2b^3?

vladimir1956 [14]

7a^2b + 10a^2b^2 + 14a^2b^3 =
a^2b (7 + 10b + 14b^2) <==

3 0

4 years ago

Read 2 more answers

I NEED THE ANSWER TO THIS ASAP FOR A TEST PLEASE!!!!!!!!

Varvara68 [4.7K]

Answer:

192

Step-by-step explanation:

you really should do your own test

A = lwh

A = 6 x 4 x 5 + 3

A = 6 x 4 x 8

A = 192

7 0

3 years ago

In an EFE​ Matrix, how much more important is a weight of 0.20 than a weight of​ 0.15? A. ​50% B. ​33% C. ​25% D. ​.05% E. ​5.0%

In an EFE Matrix, how much more important is a weight of 0.20 than a weight of 0.15? A. 50% B. 33% C. 25% D. .05% E. 5.0%