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sergij07 [2.7K]
2 years ago
12

A typical lap around a track in the United States has a length of 440 yards.How many laps would need to be completed to run a mi

le
Mathematics
2 answers:
docker41 [41]2 years ago
7 0
There are 5,280 feet in a mile so divided by three gets how many yards: 1760. And 1,716 divided by 4 is 440. So Four (4) laps.
Gekata [30.6K]2 years ago
6 0

Answer:

Hence, 4 laps would be need to  be completed to run a mile.

Step-by-step explanation:

We know that there are 1760 yards in one mile.

i.e. 1 mile=1760 yards.

As it is given that a typical lap around a track in the United States has a length of 440 yards.

i.e. we are given a number of yards one covers in 1 lap.

That means 440 yards=1 lap

so 1 yard=1/440 lap

⇒  1760 yard=1760/440 lap

⇒   1760 yard=4 lap.

Hence, 4 laps would be need to  be completed to run a mile.

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A stretch will make the figure _________________ than the original figure WHILE a shrink will make the figure __________ the ori
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A stretch will make the figure longer or wider than the original figure while a shrink will make the figure smaller then the original figure.

Step-by-step explanation:

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a real estate company sells 8 houses per month write an equation to find the total number of houses h sold in any number of mont
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H=8x,so a real estates will sell 120 houses in 15 months. Hope it help!
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A person sitting in the top row of the bleachers at a sporting event drops a pair of sunglasses from a height of 24 feet. The fu
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The 2003 Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United Sta
soldi70 [24.7K]

Answer:

a. P(x=0)=0.2967

b. P(x=1)=0.4444

c. P(x=2)=0.2219

d. P(x=3)=0.0369

Step-by-step explanation:

The variable X: "number of meals that exceed $50" can be modeled as a binomial random variable, with n=3 (the total number of meals) and p=0.333 (the probability that the chosen restaurant charges mor thena $50).

The probabilty p can be calculated dividing the amount of restaurants that are expected to charge more than $50 (5 restaurants)  by the total amount of restaurants from where we can pick (15 restaurants):

p=\dfrac{5}{15}=0.333

Then, we can model the probability that k meals cost more than $50 as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{3}{k} 0.333^{k} 0.667^{3-k}\\\\\\

a. We have to calculate P(x=0)

P(x=0) = \dbinom{3}{0} p^{0}(1-p)^{3}=1*1*0.2967=0.2967\\\\\\

b. We have to calculate P(x=1)

P(x=1) = \dbinom{3}{1} p^{1}(1-p)^{2}=3*0.333*0.4449=0.4444\\\\\\

c. We have to calcualte P(x=2)

P(x=2) = \dbinom{3}{2} p^{2}(1-p)^{1}=3*0.1109*0.667=0.2219\\\\\\

d. We have to calculate P(x=3)

P(x=3) = \dbinom{3}{3} p^{3}(1-p)^{0}=1*0.0369*1=0.0369\\\\\\

6 0
2 years ago
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