Ask question

Ask question

All categories

3 years ago

13

An edging was placed around a circular garden. How long was the edging if the radius of the garden was 3m. Use TT = 3.14

1 answer:

Tom [10]3 years ago

4 0

Edging is 18.84m. to find the edging you must do the radis times 2, then that answer by pi (3.14). So you do 3x2=6, then 6x3.14=18.84m

You might be interested in

Angle 1 and angle 2 are vertical angles. If m angle 1 = (8x+12)° and m angle 2 = (3x+37)° find the measure for each angle.

laila [671]

Vertically opposite angles are always equal.

Given angles are (8x+12)° and (3x+37)°

=> (8x+12)° = (3x + 37)°

=> 8x - 3x = 37-12

=> 5x = 25

=> x = 25/5 = 5°

Angle 1 = (8x+12)° = (8(5)+12) = 40+12 = 52°

Angle 2 = (3x + 37)° = (3(5)+37)° = 15+37 = 52°

8 0

2 years ago

Gala2k [10]

Answer:

3-11n>21-5n

Step-by-step explanation:

It seems right, so hopefully it is.

7 0

3 years ago

The circumference of a circle is 18.84 inches. what is the circle's radius

vivado [14]

Answer:

Circumference : 18.84 inches

Radius: 3 inches

6 0

3 years ago

There is a 172 cm border triangle. One side is 54 cm long. The length of the other two sides must be in a ratio of 5:8. What are

Licemer1 [7]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

4 0

3 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

Read 2 more answers