Samantha walked 10 miles in 5 hours. What was her walking rate in miles per hour?

An increase in walking has been shown to contribute to a healthier life-style. A sedentary American takes an average of 5000 ste

Zielflug [23.3K]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 5000

For the alternative hypothesis,

H1: µ > 5000

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 5000

x = 5430

σ = 600

n = 40

z = (5430 - 5000)/(600/√40) = 4.53

Looking at the normal distribution table, the probability corresponding to the z score is < 0.0001

Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, at a 5% level of significance, it can be concluded that they walked more than the mean number of 5000 steps per day.

6 0

1 year ago

What is the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)? Round to the nearest hundredth

uysha [10]

Answer:

The perimeter of the trapezoid is $38.25\ units$

Step-by-step explanation:

we know that

The perimeter of the trapezoid is the sum of its four side lengths

so

In this problem

$P=QR+RS+ST+QT$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$Q(8, 8), R(14, 16), S(20, 16),T(22, 8)$

step 1

Find the distance QR

$Q(8, 8), R(14, 16)$

substitute the values in the formula

$d=\sqrt{(16-8)^{2}+(14-8)^{2}}$

$d=\sqrt{(8)^{2}+(6)^{2}}$

$d=\sqrt{100}$

$QR=10\ units$

step 2

Find the distance RS

$R(14, 16), S(20, 16)$

substitute the values in the formula

$d=\sqrt{(16-16)^{2}+(20-14)^{2}}$

$d=\sqrt{(0)^{2}+(6)^{2}}$

$d=\sqrt{36}$

$RS=6\ units$

step 3

Find the distance ST

$S(20, 16),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-16)^{2}+(22-20)^{2}}$

$d=\sqrt{(-8)^{2}+(2)^{2}}$

$d=\sqrt{68}$

$ST=8.25\ units$

step 4

Find the distance QT

$Q(8, 8),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-8)^{2}+(22-8)^{2}}$

$d=\sqrt{(0)^{2}+(14)^{2}}$

$d=\sqrt{196}$

$QT=14\ units$

step 5

Find the perimeter

$P=10+6+8.25+14=38.25\ units$

6 0

2 years ago

Read 2 more answers

The mean score of a competency test is 82, with a standard deviation of 2. Between what two values do about 99.7% of the values

goldenfox [79]

99.7% encompasses about 3 standard deviations either side of the mean.

82 ±3*2 = (76, 88)

About 99.7% of the values lie between 76 and 88.

5 0

1 year ago

Read 2 more answers

a homeowner plans to hang wallpaper on one wall of a bedroom that is 10 feet long. if a strip of wallpaper is 20 inches wide and

love history [14]

1 ft = 12 inches....so 10 ft = (10 * 12) = 120 inches

each strip of wall paper is 20 inches...

so 120/20 = 6 strips of wall paper are required <==

4 0

1 year ago

A law school administrator was interested in whether a student's score on the entrance exam can be used to predict a student's g

frutty [35]

Answer:

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the $r^2$ who represent the determination coefficient and we got:

$r^2 = 0.934^2= 0.872$

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the $r^2$ who represent the determination coefficient and we got:

$r^2 = 0.934^2= 0.872$

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

8 0

1 year ago