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2 years ago

Use the information below to answer 1–3.

Mathematics

1 answer:

lilavasa [31]2 years ago

8 0

Answer:

The probability the 1st player does not get a blue chracter is 9/14.

Step-by-step explanation:

So there is a total of 14 characters.

9 of them are not blue

9/14 = .642857 (~64%, not the answer since it's not asking for percents)

The probability the 1st player does not get a blue chracter is 9/14.

You might be interested in

Carissa counts by 5s from 32 to 62. Which number does carissa count?

Anna35 [415]

Answer:

32, 37, 42, 47, 52, 57, 62.

Step-by-step explanation:

It is given that Carissa counts by 5s from 32 to 62.

It means, we have to add 5 in the number to get the next number.

Add 5 in 32.

$32+5=37$

Now, add 5 in 37.

$37+5=42$

Similarly, continue the process.

$42+5=47$

$47+5=52$

$52+5=57$

$57+5=62$

Therefore, the numbers counted by carissa are 32, 37, 42, 47, 52, 57, 62.

4 0

3 years ago

7. What is a fair price to pay for the spinner game shown here?

andreyandreev [35.5K]

Land on a and win $1 and land on a b and win $2 and land in

8 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

I need help with finding out both of the answers to these questions.

Semmy [17]

Answer:

6). unknown side is 6.23m

7). angle is 26.58 degrees

Step-by-step explanation:

you must use the Pythagoras theorem formula to find the unknown side if you have only two sides

7 0

3 years ago

Read 2 more answers

How many times larger is 12 x 104 than 4 x 102?

GenaCL600 [577]

Answer:

3.05

Step-by-step explanation:

12x104= 1,248

4x102= 408

1248/408-3.05

7 0

3 years ago