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77julia77 [94]
2 years ago
8

A circular track is 1000 yards in circumference. Cyclists A, B, and C start at the same place and time, and race around the trac

k at the following rates per minute: A at 700 yards, B at 800 yards, and C at 900 yards. What is the least number of minutes it mus take for all three to be together again?
Mathematics
1 answer:
Lera25 [3.4K]2 years ago
6 0

Answer:

B gains 100 yards every minute. So, after 10 minutes, B has lapped A once.

C gains 100 yards on B every minute, so after 10 minutes C has lapped B.

Thus, after 10 minutes, A has gone 7 laps, B has gone 8 laps, C has gone 9 laps, and all are even again.

You might be interested in
An arena receives $1,700 per event for 12 concerts. If the costs for this sponsorship total $14,000, what is the profit margin f
miss Akunina [59]

Answer:

total profit margin for all 12 concert is $6,400.

profit margin for all one concert is  $533.33

Step-by-step explanation:

Amount of money received for one event = $1700

Total no of concert = 12

Total money received for 12 event = Amount of money received for one event * Total no of concert

Total money received for 12 event  = $1700 *12 = $20,400

Total cost of sponsorship = $14,000

Profit margin is the difference money invested and money earned.

here total investment is $14,000

Total money earned is  $20,400

Therefore profit margin = money earned - total investment

= $20,400 -  $14,000 = $6,400

Therefore total profit margin for all 12 concert is $6,400.

However to calculate profit margin for one concert we can simply divide the total profit margin for all 12 concert by no of concert (i.e 12)

profit margin for one concert = total profit margin for all 12 concert/total no of concert  = $6,400/12 = $533.33

profit margin for all one concert is  $533.33

6 0
2 years ago
Read the numbers and decide what the next number should be. 256 196 144 100 64 36 16
Tatiana [17]
340? TBD add the two numbers.
4 0
2 years ago
Read 2 more answers
Solve the following inequality: –1 + 6(–1 – 3x) > –39 – 2x.    
insens350 [35]
<span>–1 + 6(–1 – 3x) > –39 – 2x. 
</span>-1-6-18x>-39-2x
-7-18x>-39-2x
-18x>-32-2x
-16x>-32
x<2


B. x<2
7 0
2 years ago
Read 2 more answers
Jenna gets an iTunes gift card for her birthday and decides to buy some games and some music the music she likes costs $1.99 per
alexandr1967 [171]

Answer: 4 songs and 11 games

Step-by-step explanation:

Sorry if I’m wrong!

3 0
1 year ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390

<u>For X=8</u>

This happens when everybody is late

P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147

8 0
2 years ago
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