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Leto [7]
3 years ago
11

If it takes you 40 minutes to go 20 miles downstream and then 60 minutes on the way back, what is the speed of the current?

Mathematics
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

Speed of Current = 5 miles per hour

Step-by-step explanation:

We know distance formula,

D = RT

Where

D is distance

R is rate

T is time

If we let speed of boat (assume) to be "x" and speed of current to be "c"

Then downstream rate is (with current) = x + c

Upstream rate is (against current) = x - c

40 mins to go 20 miles downstream, that means:

D = RT

20 = (x + c)(40)

and

60 minutes to go upstream, 20 miles, that means:

D = RT

20 = (x - c)(60)

Simplifying first equation:

40x + 40c = 20

Simplifying second equation:

60x - 60c = 20

Multiplying first equation by 60, we get:

60 * [40x + 40c = 20] = 2400x + 2400c = 1200

Multiplying second equation by 40, we get:

40 * [60x - 60c = 20] = 2400x - 2400c = 800

Now we add up both these equations:

2400x + 2400c = 1200

2400x - 2400c = 800

----------------------------------

4800x = 2000

x = 2000/4800 = 5/12

We need speed of current, that is "c", so we plug in the value of x into first equation and solve for c:

40x + 40c = 20\\40(\frac{5}{12}) + 40c = 20\\\frac{50}{3}+40c=20\\40c=20-\frac{50}{3}\\40c=\frac{10}{3}\\c=\frac{1}{12}

Speed of Current = 1/12 miles per minute

Since there is 60 minutes in an hours, that would be:

(1/12) * 60 = 5 miles per hour

Speed of Current = 5 miles per hour

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Answer:

A) Option B is correct.

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Hₐ: μ₁ - μ₂ < 0

B) t = -2.502

p-value = 0.0112

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Step-by-step explanation:

The complete Question is presented in the two attached images to this answer.

A) To perform this test we first define the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.

Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂

The null hypothesis is represented as

H₀: μ = 0 or μ₁ = μ₂

The alternative hypothesis is represented as

Hₐ: μ < 0 or μ₁ - μ₂ < 0

B) So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₁ - μ₂))/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average compression strength of strawberry drink = 537

n₁ = sample size of the sample of strawberry drink in cans surveyed = 10

s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22

μ₂ = average compression strength of cola = 559

n₂ = sample size of the sample of cola in cans surveyed = 10

s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17

σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792

We will use the t-distribution as no information on population standard deviation is provided

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checking the tables for the p-value of this t-statistic

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C) The interpretation of p-values is that

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0.0112 < 0.05

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D) The necessary conditions required before a t-test is deemed valid include.

- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.

- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.

Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.

Hope this Helps!!!

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