All categories

3 years ago

53% of the 1000 students at Barnett High are girls. How many boys are there in the school?

Mathematics

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

470

Step-by-step explanation:

1000 x 0.53 = 530

1000 - 530 = 470

You might be interested in

Jessica used 2/3 yard of fabric to make a scarf. Can she make 2 of these scarves with

mr Goodwill [35]

Yes because
1 3/4 is 7/2 and 2/3+2/3=4/3

So yes he can
Hope this helps :-P

5 0

3 years ago

Read 2 more answers

Answer this 4(−12)(−3)

KatRina [158]

Answer:

144

Step-by-step explanation:

3 0

4 years ago

Last year, the school library had a total of x books. Over the summer, the library acquired another 42 books and now has a total

natta225 [31]

B.x + 42 = 2,402 letter b

4 0

3 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

3 years ago

An automobile insurance company divides customers into three categories: good risks, medium risks, and poor risks. Assume that o

Leokris [45]

Answer:

0.60

Step-by-step explanation:

Probability that the customer is not a poor risk = 1 - probability that the customer is a poor risk

Firstly, let’s calculate the probability of being a poor risk.

From the given data the number of poor risks = 14229-7362-1190 = 5677

So the probability of being a poor risk = 5677/14229 = 0.399

Thus, the probability that the customer is not a poor risk = 1-0.399 = 0.601 which to 2 decimal places = 0.60

6 0

3 years ago