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2 years ago

It takes Andre 4 minutes to swim 5 laps.

Mathematics

2 answers:

soldier1979 [14.2K]2 years ago

6 0

It takes Andre 1.25 minutes to swim a lap.

lbvjy [14]2 years ago

5 0

Answer:

4/5

Step-by-step explanation:

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Is a measure of 26 inches "far away" from a mean of 16 inches? As someone with knowledge of statistics, you answer "it depen

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Answer: Hello!

In a normal distribution, between the mean and the mean plus the standar deviation, there is a 34.1% of the data set, between the mean plus the standar deviation, and the mean between two times the standard deviation, there is a 16.2% of the data set, and so on.

If our mean is 16 inches, and the measure is 26 inches, then the difference is 10 inches between them.

a) if the standar deviation is 2 inches, then you are 10/2 = 5 standar deviations from the mean.

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c) Now the standar deviation is 7, so now 26 is in the range between 1 standar deviation and 2 standar deviations away from the mean.

Then this you have a 16% of the data, then in this case, 26 inches is not far away from the mean.

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Luke bikes for 2.58 miles and takes a car for 6.93 miles each morning to reach his school. How long is Luke's journey to school?

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3 years ago

The owner of a flower shop sells18 roses for $90.00. What is the cost in dollars per rose?

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3 years ago

Read 2 more answers

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago