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Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus

quester [9]

Hi!

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$ ≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.

5 0

3 years ago

Solve.<br> x<br> __<br> 2 = 7<br> ----------------------------------------<br> PLEASE HELP!

galina1969 [7]

$\frac{x}{2} = 7\\\\\frac{x}{2} \times 2 = 7\times2\\\\\boxed{x=14}$

5 0

3 years ago

A rectangular parking lot has a perimeter of 820 ft. The area of the parking lot measures 42,000 ft2. What is a dimension of the

dalvyx [7]

To solve the problem we must know about quadratic equations.

<h2>Quadratic Equation</h2>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

The breadth of the rectangle is 200 ft, while the length is 210 ft.

<h2>Explanation</h2>

Given to us

Area of the parking lot = 42,000 ft²
Perimeter of the parking lot = 820 ft

<h3>Area of the parking lot</h3>

Area of the parking lot = Area of the rectangle

42,000 ft² = Length x Breadth

Solving for L,

$42,000 = L \times B\\\\
L = \dfrac{42,000}{B}$

<h3>Perimeter of the parking lot</h3>

Perimeter of the parking lot = Perimeter of the rectangle

820 ft. = 2(Length + Breadth)

820 ft. = 2(L+ B)

$2(L+ B) = 820\\\\
(L+B) = \dfrac{820}{2}\\\\
(L+B) = 410$

Substituting the value of L,

$(L+B) = 410\\\\
(\dfrac{42,000}{B}) +B = 410\\\\
42000 + B^2 = 410B\\\\
B^2 -410B +42000 = 0$

<h3>Quadratic Expression</h3>

Solving the quadratic Expression,

$B^2 -410B +42000 = 0\\\\
(B-210)(B-200)=0$

Equation the factors against zero,

B-210=0

B = 210

B-200=0

B = 200

Hence, the breadth of the rectangle is 200ft, while the length is 210 ft.

Learn more about Quadratic Expression:

brainly.com/question/10025464

4 0

3 years ago

Which is the equation of a line that is perpendicular to the line represented by y=7/6x+4

Mumz [18]

B because to be perpendicular the slope had to be opposite of the original slope but the same b. But parallel would be the exact same slope with a different b

5 0

3 years ago

What temperature is 23 degrees above freezing in fahrenheit

jek_recluse [69]

The temperature that is 23 degrees above freezing is 55 because the freezing level is 32 and you add 23 to that

7 0

3 years ago