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worty [1.4K]
3 years ago
6

A pendulum swings through an angle of 20 degrees each second. If the pendulum is 40 inches long how far does the tip move each s

econd? does r= 20 in. or 40 in
Mathematics
1 answer:
Olin [163]3 years ago
3 0

If the pendulum is 40 inches long, its tip moves 13.96” each second. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Determine whether angles with measures 47 ° and 53 ° are complementary. Explain why or why not.
Stells [14]

Answer: No

Step-by-step explanation: Complementary angles are those that add up to 90 degrees, 47 + 53 equals to 100.

6 0
3 years ago
1. Charlene wants to center a rectangular pool in her backyard so that the edges of the pool are an equal distance from the edge
Inga [223]

First, we have to calculate the length of the sides of the pool, we are told that the scale factor of the backyard to the pool size equals 1/2, then we can find the length of the sides of the pool by multiplying the lengths of the sides of the backyard by 1/2, like this:

length of the pool = length of the yard * 1/2

width of the pool = width of the yard * 1/2

By replacing the 60 m for the length of the yard and 50 m for the width, we get:

length of the pool = 60 * 1/2 = 30

width of the pool = 50 * 1/2 = 25

Let's call x1 to the distance from the base of the pool to the bottom side of the yard and x2 to the distance from the top side of the pool to the top side of the yard, then we can formulate the following equation:

width of the yard = width of the pool + x1 + x2

Since we want the edges to be at an equal distance, x1 and x2 are the same, then we can rewrite them as x:

width of the yard = width of the pool + x + x

width of the yard = width of the pool + 2x

Replacing the known values:

60 = 30 + 2x

From this equation, we can solve for x to get:

60 - 30 = 30 - 30 + 2x

30 = 2x

30/2 = 2x/2

15 = x

x = 15

Now, let's call y1 to the distance from the right side of the corresponding side of the yard and y2 to the distance from the left side of the pool to the left side of the yard, with this, we can formulate the following equation:

length of the yard = length of the pool + y1 + y2

Since we want the edges to be at an equal distance, y1 and y2 are the same, then we can rewrite them as y:

length of the yard = length of the pool + y + y

length of the yard = length of the pool + 2y

Replacing the known values:

50 = 25 + 2y

50 - 25 = 25 - 25 + 2y

25 = 2y

25/2 = 2y/2

12.5 = y

y = 12.5

Now, we know that the pool must be at a distance of 15 m from the horizontal sides of the pool to the horizontal sides of the yard and that it must be at a distance of 12.5 m from the vertical sides of the pool to the vertical sides of the yards.

Here is a figure that depicts the results:

5 0
1 year ago
Saul in third in line. Enid is last in line. There are 3 children between them. What position is Enid in line?
VLD [36.1K]
Enid is seventh in line
8 0
4 years ago
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Aleonysh [2.5K]
The answer would be D yeahhhhhh
3 0
3 years ago
Read 2 more answers
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