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loris [4]
3 years ago
9

The perimeter of a rectangle is 96 inches. The length of the rectangle is three times its width. Find the dimensions of the rect

angle.
Mathematics
1 answer:
Oksanka [162]3 years ago
4 0

Hey there!!

Perimeter of a rectangle =2 ( length + width )

Let's take the measure of the width as ' x '

Then, the length would be ' 3x '

Hence,

... 2 ( x + 3x ) = 96

... 2 ( 4x ) = 96

... 8x = 96

... x = 96 / 8

... x = 12

Width = 12 inches

And, length would be 36 inches.

Hope my answer helps!!

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DATA ANALYSIS AND STATISTICSProbabilities involving two rolls of a dieAn ordinary (falr) die is a cube with the numbers 1 throug
Zepler [3.9K]

Solution

The table below is the required sample space of the to fair die

From the above table

The sample space contain 36 outcomes

Event A: The sum is greater than 9

we will look at the table and count all the elements that are greater than 9

There are 6 elements (they are 10, 10, 10, 11, 11, 12 from the table)

The probability for event A will be

\begin{gathered} p(A)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(A)=\frac{6}{36} \\ p(A)=\frac{1}{6} \end{gathered}

P(A) = 1/6

Event B: The sum is an even number.

We will look at the table and count the number of elements that are even

There are 18 elements (notice that there are 3 even number on each of the 6 rows of the table)

The probability for event B will be

\begin{gathered} p(B)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(B)=\frac{18}{36} \\ p(B)=\frac{1}{2} \end{gathered}

p(B) = 1/2

5 0
11 months ago
The amount of time it takes Isabella to wait for the bus is continuous and uniformly distributed between 3 minutes and 17 minute
Arturiano [62]

Answer:

P(X>5) = 0.857

Step-by-step explanation:

Let X \sim uniform(3.17)

f(x) = \dfrac{1}{17-3}   ; \ \ \ 3 \le x \le 17

The required probability that it will take Isabella more than 5 minutes to wait for the bus can be computed as:

P(X > 5) =  \int ^{17}_{5} f(x) \ dx

P(X > 5) =  \int ^{17}_{5} \dfrac{1}{17-3} \ dx

P(X > 5) =\dfrac{1}{14}  \Big [x \Big ] ^{17}_{5}

P(X > 5) =\dfrac{1}{14} [17-5]

P(X > 5) =\dfrac{12}{14}

P(X>5) = 0.857

6 0
2 years ago
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