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Alborosie
3 years ago
11

the bleachers at the football game are 7/8 full, and 1/2 of the fans in the bleachers are rooting for the home team. what fracti

on of the bleachers are filled with home-team fans?
Mathematics
1 answer:
wolverine [178]3 years ago
5 0
First you have to find a common denominator which is 8.
1/2 =4/8
7/8 - 4/8 = 3/8
So the answer is 3/8.
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What is the sum of 10/12and 11/12? Make sure to show all work to receive full credit!
sergiy2304 [10]

Answer:

1\frac{3}{4}

Step-by-step explanation:

\frac{10}{12} +\frac{11}{12} \\\\\frac{21}{12} \\\\1\frac{9}{12} \\\\1\frac{3}{4}

7 0
3 years ago
More than 450 students traveled to a state park for a field trip. The school allowed 6students to travel by car, and the rest tr
ki77a [65]

Answer:


Step-by-step explanation:

Let b= number of people per bus

 

10b + 5 > 450

 

10b > 445

b > 44.5

Minimum per bus is 45 students

6 0
3 years ago
Solve x2 + 8x + 22 = 0 by completing the square.
Liula [17]

Answer:

A)

Step-by-step explanation:

the solution of a squared equation is

x = (-b ± sqrt(b² - 4ac)) / (2a)

in our case

a = 1

b = 8

c = 22

x = (-8 ± sqrt(64 - 88))/2 = (-8 ± sqrt(-24))/2 =

= (-8 ± sqrt(4×-6))/2 = (-8 ± 2×sqrt(-6))/2 =

= -4 ± sqrt(-6) = -4 ± i×sqrt(6)

6 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
draw diagnose AC and BD on parallelogram ABCD. Make a point where the diagonals intersect, and label it E. Measure and record th
VARVARA [1.3K]
I need the answer to
3 0
3 years ago
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