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3 years ago

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The picture shows the footprints of a man walking. The pace length P is the distance between the rear of two consecutive footpri

nts.
For men, the formula, n P =140, gives an approximate relationship between n and P where,
n = number of steps per minute, and
P = pace length in metres.
b
Bernard knows his pace length is 0.80 meters. The formula applies to Bernard’s walking. Calculate Bernard’s walking speed in meters per minute and in kilometers per hour. Show your working out.

1 answer:

Rina8888 [55]3 years ago

7 0

Bernard’s walking speed is 140 meters per minute

Bernard’s walking speed is 8.4 kilometers per hour

Step-by-step explanation:

The pace length P is the distance between the rear of two consecutive footprints

The formula, n P = 140, gives an approximate relationship between n and P where,

n = number of steps per minute
P = pace length in meters
Bernard knows his pace length is 0.80 meters
The formula applies to Bernard’s walking

We need to calculate Bernard’s walking speed in meters per minute and in kilometers per hour

∵ n = number of steps per minute

∴ n unit is number / minute

∵ P = pace length in meters

∴ P unit is meter

- That means n P unit is meters/minute

∴ n P represents the speed in meters per second

∵ The formula applies to Bernard’s walking

∵ His pace length is 0.80 meters

∵ n P = 140

∴ n(0.8) = 140

- Divide both sides by 0.8

∴ n = 175

- That means he moves 175 steps per minute

∵ The distance he walks in minute = 175 × 0.8 = 140 meters/minute

∴ His speed is 140 meters per minute

Bernard’s walking speed is 140 meters per minute

∵ 1 km = 1000 m

∵ 1 hour = 60 minutes

- Divide the meters by 1000 and the minute by 60 to change

from meters per minute to kilometers per hour

∴ His walking speed = $\frac{140}{1000}$ ÷ $\frac{1}{60}$

- Change ÷ to × and reciprocal the fraction after the division sign

∴ His walking speed = $\frac{140}{1000}$ × $\frac{60}{1}$ = 8.4 km/h

Bernard’s walking speed is 8.4 kilometers per hour

Learn more:

You can learn more about the speed in brainly.com/question/5461619

#LearnwithBrainly

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We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys

marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation

$\bar X=19.2$ represent the battery life sample mean

$\sigma=2.5$ represent the population standard deviation

$n=25$ sample size

$\mu_o =18$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:

Null hypothesis: $\mu = 18$

Alternative hypothesis: $\mu \neq 18$

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis: $\mu \leq 18$

Alternative hypothesis: $\mu > 18$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test for parts I and II, the p value would be:

$p_v =2*P(t_{(24)}>2.4)=0.0245$

And for part III since we have a one right tailed test the p value is:

$p_v =P(t_{(24)}>2.4)=0.0122$

5) Conclusion

I. This finding is significant for a two-tailed test at .05.

Since the $p_v$ . We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0

3 years ago

Order these units from smallest to largest:

Veronika [31]

The units are ordered smallest to largest from top to bottom:
Picometer (pm) = 1 x 10⁻¹² meters
Nanometer (nm) = 1 x 10⁻⁹ meters
Micrometer (um) = 1 x 10⁻⁶ meters
Millimeter (mm) = 1 x 10⁻³ meters
Centimeter (cm) = 1 x 10⁻² meters
Decimeter (dm) = 1 x 10⁻¹ meters
Meters (m) = 1 meter
Kilometers (Km) = 1 x 10³ meters

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3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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IRINA_888 [86]

Answer: C

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