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REY [17]
3 years ago
12

Given V=1/3(3.14)r^2h solve for r

Mathematics
1 answer:
Gnoma [55]3 years ago
5 0

The answer is r = √ v/1.046667h   (v over 1.046667h)

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1 foot = 12 inches<br> 1 inch = 2.5 centimeters<br> change 5 feet and 10 inches to centimeters
jolli1 [7]

Answer:

The answer should be 177.8 cm.

Step-by-step explanation:

4 0
3 years ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
4 years ago
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3 0
4 years ago
Are rectangular garden has a perimeter of 280 feet the north side of the garden is 90 feet what is the length of the inside of t
attashe74 [19]

The length of the inside of the garden is<u> 50 feet.</u>

What is perimeter ?

A perimeter is a closed path that covers, surrounds, or outlines a two-dimensional shape or length. The circumference of a circle or an ellipse is its perimeter.

There are various practical applications for calculating the perimeter. A calculated perimeter is the length of fencing needed to completely encircle a yard or garden. The perimeter (circumference) of a wheel/circle describes how far it will roll in one revolution. Similarly, the amount of string wrapped around a spool is proportional to the spool's perimeter; if the string length were correct, it would equal the perimeter.

Given :

The rectangular garden has a perimeter of 280 feet

breadth = 90 feet

Let the length be x.

Therefore according to the question,

2( 90 + x ) = 280

=> 90+ x = 140

=> x = 50

To learn more about perimeter click on the link below:

brainly.com/question/397857

#SPJ10

6 0
2 years ago
Help me please! I don’t quite get this <br> 10 pts!
storchak [24]

Step-by-step explanation:

Mathematical explanations are explanations in which mathematics plays a crucial role. There is a rich historical tradition of debates about mathematical explanation, dating back to Aristotle. One central strand of this tradition considers the significance of explanatory goals for pure mathematics.

4 0
3 years ago
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