Use the values on the number line to find the sampling error.

A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard

Sauron [17]

Answer:

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $393.50

Standard deviation r = $50.30

Number of samples n = 25

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

$393.50+/-1.96($50.30/√25)

$393.50+/-1.96($10.06)

$393.50+/-$19.7176

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

5 0

2 years ago

Find the semi perimeter of a triangle with sides 9cm 12 cm and 30 cm

alexgriva [62]

Answer:

Step-by-step explanation:

$= \frac{9 + 12 + 30}{2}cm \\ = \frac{51}{2}cm \\ 25.5cm$

7 0

2 years ago

15x²y³z÷3xy² simplify

umka21 [38]

Answer:

$5xyz$

Step-by-step explanation:

$\frac{15x^2y^3z}{3xy^2}$

Step 1: Divide the numbers

$\frac{15}{3}=5\\\\=\frac{5x^2y^3z}{xy^2}$

Step 2: Simplify

$\frac{x^2}{x}\\\\= x\\\\=\frac{5xy^3z}{y^2}$

Step 3: Simplify

$\frac{y^3}{y^2} = y\\\\=5xyz$

Therefore, the simplified answer is $=5xyz$

7 0

1 year ago

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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

2 years ago

Match each three-dimensional figure to its volume based on the given dimensions. (Assume π = 3.14.)

LekaFEV [45]

Answer:

The volume of the cylinder is 150.72 cm³ ⇒ last answer

The volume of the cone is 314 cm³ ⇒ 1st answer

The volume of the pyramid is 160 cm³ ⇒ 2nd answer

The volume of the pyramid is 48 cm³ ⇒ 3rd answer

Step-by-step explanation:

* Lets revise the volumes of some shapes

- The volume of the cylinder of radius r and height h is:

V = π r² h

- The volume of the cone of radius r and height h is:

V = 1/3 π r² h

- The volume of the pyramid is:

V = 1/3 × its base area × its height

* Lets solve the problem

# A cylinder with radius 4 cm and height 3 cm

∵ V = π r² h

∵ π = 3.14

∵ r = 4 cm , h = 3 cm

∴ v = 3.14 (4)² (3) = 150.72 cm³

* The volume of the cylinder is 150.72 cm³

# A cone with radius 5 cm and height 12 cm

∵ V = 1/3 π r² h

∵ π = 3.14

∵ r = 5 cm , h = 12 cm

∴ V = 1/3 (3.14) (5)² (12) = 314 cm³

* The volume of the cone is 314 cm³

# A pyramid with base area 16 cm² and height 30 cm

∵ V = 1/3 × its base area × its height

∵ The area of the base is 16 cm²

∵ The height = 30 cm

∴ V = 1/3 (16) (30) = 160 cm³

* The volume of the pyramid is 160 cm³

# A pyramid with square base of length 3 cm and height 16 cm

∵ V = 1/3 × its base area × its height

∵ The area of the square = s²

∵ The area of the base = 3² = 9 cm²

∵ The height = 16 cm

∴ V = 1/3 (9) (16) = 48 cm³

* The volume of the pyramid is 48 cm³

3 0

2 years ago

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