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Komok [63]
2 years ago
6

In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard dev

iation of 5.4 minutes. It is assumed that this is a normally distributed variable. Find the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.
a. 0.38
b. 0.69
c. 0.99
d. 0.31
Mathematics
1 answer:
Paha777 [63]2 years ago
3 0

Answer:

0.3114

Option d is right

Step-by-step explanation:

Let X be the time spent on a treadmill in the health club

Given that  research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes

Also given that X is normal

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.

= P(30

round off to two decimals tog et

the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill is 0.31

Hence option d is right

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yKpoI14uk [10]

Answer:

60

Step-by-step explanation:

6 0
2 years ago
Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that
Vadim26 [7]
So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\

\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format

8 0
2 years ago
Read 2 more answers
Is AB parallel to Bo? Explain.
Svetradugi [14.3K]

Answer:

AB parallel to CD because both lines have a slope of

of 4/3

Step-by-step explanation:

The question is not complete, there is no graph.

A graph for the question is attached below.

From the image attached below, line 1 passes through points A = (-3, -3) and point B = (0, 1) while line 2 passes through point C = (0, -5) and point D = (3, -1).

Two parallel are said to be parallel if the have the same slope. The slope of a line passing through points:

(x_1,y_1)\ asnd\ (x_2,y_2).\ The\ slope \ is\ given\ as:\\\\Slope(m)=\frac{y_2-y_1}{x_2-x_1}

Line 1 passes through points A = (-3, -3) and point B = (0, 1), the slope of line 1 is:

Slope(m)=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-3)}{0-(-3)}=\frac{1+3}{0+3}=\frac{4}{3}

Line 2 passes through point C = (0, -5) and point D = (3, -1). the slope of line 2 is:

Slope(m)=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-(-5)}{3-0}=\frac{-1+5}{3}=\frac{4}{3}

Therefore AB parallel to CD because both lines have a slope of

of 4/3

7 0
2 years ago
Ann's car can travel 228 miles on 6 gallons of gas. a. Write an equation to represent the distance, y, in miles Ann's car can tr
Murrr4er [49]

Answer:

288 gallons ÷ 6 miles = 48 miles per gallon  

48 miles per gallon • 7 miles = 366 miles  

Step-by-step explanation:

7 0
1 year ago
We suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in y
ANTONII [103]

Answer:

D) a chi square test for independence.

Step-by-step explanation:

Given that we  suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums.

We are to check whether two variables insurance premiums and driving experience are associated.

Two categorical variables are compared for different ages and insurance premiums.

Hence a proper test would be

D) a chi square test for independence.

8 0
1 year ago
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