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2 years ago

I can buy 4 basketballs for $44. At that same rate, how much would 8 basketballs cost

Mathematics

2 answers:

lorasvet [3.4K]2 years ago

7 0

Answer

Step-by-step explanation:

4 bb:$44

If you do eight divided by four you'll get two, so then you'll have to divide two on 44 which is 22.

Advocard [28]2 years ago

4 0

Answer:

$88.00

Step-by-step explanation:

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Period 1 has 15 students in it and a test average of 86%. Period 2 has 21 students in it and an average of 88%. Period 3 has 12

STatiana [176]

Add all of the percents up and then divide by 3. You get 89.6%

7 0

2 years ago

Read 2 more answers

Using the greatest common factor, reduce the fraction \frac{32}{48} 48 32 to its simplest form

Vinvika [58]

Answer:

2/3

Step-by-step explanation:

32/48= 16*2/16*3= 2/3

3 0

2 years ago

The sum of 48 and itself its half and half of the hal is added to 18

hammer [34]

Answer:

150

Step-by-step explanation:

We are carrying out Addition

a) The sum of 48 and itself

= 48 + 48 = 96

b) Its half and half of the half

96 + (1/2 × 48) + (1/2 ×( 1/2 × 48)

= 96 + 24 +(1/2 × 24)

= 96 + 24 + 12

= 132

c) is added to 18

= 132 + 18

= 150

Therefore, the sum of 48 and itself, its half and half of the half is added to 18 is 150

8 0

2 years ago

The following diagram shows parallel lines cut be a transversal. What is the measure of Angle 2?

Crank

2x + 100 = 5x + 55
3x = 45
x = 15

2(15) + 100 = 130

180-130 = 50

so angle 2 is 50 degrees

3 0

2 years ago

Exercise 5.6.6: Selecting a committee of senators. About A country has two political parties, the Demonstrators and the Repudiat

krok68 [10]

Answer:

There are 4,148,350,734,528 ways

Step-by-step explanation:

We have

44 senators which are Demonstrators.
56 senators which are Repudiators.

(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?

We want to choose 5 Demonstrators and 5 Repudiators. The number of ways to do this is ${44} \choose {5}$ and $56 \choose 5$ respectively. Therefore, the number of ways to select the committee is given by:

${{44}\choose {5}} \times {{56}\choose{5}}=\frac{44!}{39!5!}\times\frac{56!}{51!5!}=\frac{44!56!}{51!39!5!5!}=\frac{44\times43\times42\times41\times40\times56\times55\times54\times53\times52}{5!5!}=\\\\=\frac{44\times43\times42\times41\times8\times56\times11\times54\times53\times52}{4!4!}= \frac{11\times43\times42\times41\times2\times56\times11\times54\times53\times52}{3!3!}=\\\\\frac{11\times43\times14\times41\times2\times56\times11\times18\times53\times52}{2!2!}=$

$11\times43\times14\times41\times28\times11\times18\times53\times52=4,148,350,734,528$

(b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

<u>If the speaker and vice speaker are chosen between all senators:</u> In this case, the answer will be

$44\times43\times56\times55=5,827,360.$

This is because there are (in the case of Demonstrators) 44 possibilities to choose an speaker and after choosing one, there would be 43 possibilities to choose a vice speaker. The same situation happens in the case of Repudiators.

<u>If the speaker and vice speaker are chosen between the committee:</u> In this case, the answer will be

$5\times4\times5\times4=400$ .

3 0

2 years ago