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Vlad1618 [11]
3 years ago
7

If i drive 439.92 miles on 15.6 gallons of gas, how many miles could i drive on 39.7 gallons of gas?

Mathematics
1 answer:
eduard3 years ago
7 0
Make it into a ratio: 439.92/15.6=x/39.7 where x is the unknown. cross multiply to get your answer of 1119.5 miles
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The average life span of a lynx is
dangina [55]

Answer:

20

Step-by-step explanation:

3/4 = 75%

15 ÷ 75% = 20

15 is 75% of 20.

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2 years ago
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The proportion of a population with a characteristic of interest is p = 0.37. What is the standard deviation of the sampling dis
lukranit [14]

The standard deviation (SD) of a sample proportion obtained when the sample size is 1,600 is equal to 0.0121.

<h3>What is a sample proportion?</h3>

A sample proportion can be defined as the proportion of individuals in a sample that have a specified characteristic or trait.

Mathematically, sample proportion can be calculated by using this formula:

\hat{p} = \frac{x}{n}

Where:

  • x represent the number of individuals with a specified characteristic.
  • n represent the total number of individuals in a sample.

In Mathematics, the standard deviation (SD) of a sample proportion obtained when the sample size is 1,600 can be calculated by using this formula:

Standard deviation (SD) = √(p(1 - p)/n)

Substituting the given parameters into the formula, we have;

Standard deviation (SD) = √(0.37(1 - 0.37)/1,600)

Standard deviation (SD) = √(0.37(0.63)/1,600)

Standard deviation (SD) = √(0.2331)/1,600)

Standard deviation (SD) = √0.0001457

Standard deviation (SD) = 0.0121.

Read more on standard deviation here: brainly.com/question/14467769

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8 0
1 year ago
207,567 to the nearest hundred thousand
Nezavi [6.7K]
The hundred thousands place is the 2. Since the place below that is 0, it rounds down.
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Evaluate -125^(-1/3)
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Answer:

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3 years ago
Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w
Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so p = 0.6

The week has 5 days, so n = 5

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so P(X = 5).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0
3 years ago
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