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Vinvika [58]
4 years ago
15

e. Is the a discrete random variable, a continuous random variable, or not a random variable? height of a randomly selected gira

ffe A. It is a discrete random variable. B. It is a continuous random variable. C. It is not a random variable.
Mathematics
2 answers:
irga5000 [103]3 years ago
5 0

Answer:

The correct answer is:

It is a continuous random variable (B)

Step-by-step explanation:

Continuous Random Variables are variables that take on a number of possibilities of values that cannot be counted. The values have infinite possibilities. In this example, the height of a Giraffe measured in meters can be an unlimited possibility if values say, 10.5m, 15.22m 12.0m etc. The possibilities are endless.

Discrete Random variables are variables that take on a number of possibility of occurrences that can be counted. For instance, if a dice is rolled, the possibilities can either be a 1, 2, 3, 4, 5 or 6. There are six values that can be gotten, nothing in-between.

mash [69]3 years ago
5 0

Huuuuuuh I do not understand

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Peter asked the students of his class their football scores and recorded the scores in the table shown below:
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Answer:

Step-by-step explanation:

Add all of the numbers together then divide it by 14=4

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A seamstress has a piece of cloth that is 3 yards long. She cuts it into shorter lengths of 16 inches each. How many of the shor
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6 3/4 pieces, but if you just want a whole number then 6
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4 years ago
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Vedmedyk [2.9K]

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3 years ago
A rectangular area is to be enclosed by a wall on one side and fencing on the other three sides. If 18 meters of fencing are use
Rashid [163]

Answer:

A = L W= 9*\frac{9}{2}=\frac{81}{2} m^2

Step-by-step explanation:

For this case we assume that the total perimeter is 18 ft, we have a wall and the two sides perpendicular to the wall measure x units each one so then the side above measure P-2x= 18-2x.

And we are interested about the maximum area.

For this case since we have a recatangular area we know that the area is given by:

A= LW

Where L is the length and W the width, if we replace from the values on the figure we got:

A(x)= x *(18-2x) = 18x -2x^2

And as we can see we have a quadratic function for the area, in order to maximize this function we can use derivates.

If we find the first derivate respect to x we got:

\frac{dA}{dx} = 18-4x=0

We set this equal to 0 in order to find the critical points and for this case we got:

18-4x=0

And if we solve for x we got:

x=\frac{18}{4}=\frac{9}{2} m

We can calculate the second derivate for A(x) and we got:

\frac{d^2 A}{dx^2}= -4

And since the second derivate is negative then the value for x would represent a maximum.

Then since we have the value for x we can solve for the other side like this:

L= 18-2x = 18-2 \frac{9}{2}= 18-9 =9m

And then since we have the two values we can find the maximum area like this:

A = L W= 9*\frac{9}{2}=\frac{81}{2} m^2

8 0
3 years ago
47% of 8.4 million .....
FrozenT [24]

Answer:

39,48,000

Explanation:

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