All categories

2 years ago

Question and answer options attached. Please help, thanks!

Mathematics

1 answer:

Studentka2010 [4]2 years ago

3 0

Answer:

there is no question but i think out of a,b,c and e i think b or c

Step-by-step explanation:

sorry

You might be interested in

The seats at a local baseball stadium are arranged so that each row has five more seats than the row below it. If there are 4 se

nordsb [41]

Givens:

a = 4; n = 23; d = 5

Formula

L = a + (n - 1)*d

Substitute and solve

L = 4 + (23 - 1)*5

L = 4 + 22*5

L = 4 + 110

L = 114 Answer

4 0

2 years ago

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

2 years ago

What are all the factors of 32? A 2, 4, 8, 16 B 1, 2, 16, 32 C 1, 2, 4, 8, 16, 32 D 2, 4, 6, 8, 12, 16, 32

Natasha2012 [34]

What are all the factors of 32?

answer : ( C ) 1, 2, 4, 8, 16 and 32.....

4 0

2 years ago

Read 2 more answers

A football coach set a goal of keeping opposing teams to an average of less than 200 yard. of total offense each game. After thr

faust18 [17]

(701+y)/(3+1)<200

(701+y)/4<200

701+y<800

y<99

So the most yards they can allow in the next game is 98 yards

6 0

3 years ago

What is the answer to this equation |x−1|+5=2

barxatty [35]

Answer:

No solutions

Step-by-step explanation:

Isolate the absolute value:

|x−1| + 5 = 2

Subtract 5 from both sides:

|x-1| = -3

Since an absolute value can never be equal to a negative number, there are no solutions.

6 0

2 years ago

Read 2 more answers