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xxMikexx [17]
3 years ago
11

Nina buys a small rug in the shape of a kite. The width of the rug is half its height. What are the width and area of the rug? (

height 31 inches width 9 inches)
Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Width = 20\ in

Area = 400\ in^2

Step-by-step explanation:

Given

See attachment for kite dimension

Required

Determine the width and the area

From the attachment, the height of the kite is:

Height = 31\ in + 9\ in

Height = 40\ in

The width which is half the height is:

Width = \frac{1}{2} * Height

Width = \frac{1}{2} * 40\ in

Width = 20\ in

The area is:

Area = \frac{Height * Width}{2}

Area = \frac{40\ in * 20\ in}{2}

Area = 40\ in * 10\ in

Area = 400\ in^2

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Answer:

40

Step-by-step explanation: I just guesstimated a number and divided 36 by 40 to get .9 which means 90%

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The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.
salantis [7]

a) X

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

\mu=78.1 (mean)

\sigma = 10.8 (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

X

And the probability for this to occur can be written as:

p(X

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between z=-\infty and z=Z, where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

p(X

And by looking at the z-score tables, we find that this probability is:

p(z

And so,

p(X

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

p(X>89.1)

First of all, we have to calculate the z-score corresponding to this value of X, which is:

Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

p(z>1.02) =p(z

Because the normal distribution is symmetric.

But from part b) we know that

p(z

Therefore:

p(X>89.1)=p(z>1.02)=0.1539

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

p(67.1

Or also as

p(67.1

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

p(X

p(X>89.1)=0.1539

Therefore, here we find immediately than:

p(67.1

7 0
3 years ago
1.-si el area de un rectangulo es de 144 metros cuadrados y su altura de 8 metros ¿ cual es el valor de su base?
kari74 [83]

Step-by-step explanation:

FÓRMULA:

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Answer:

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Step-by-step explanation:

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