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ExtremeBDS [4]
2 years ago
15

Part C of Jim and his dad are building a rectangular flower bed.

Mathematics
1 answer:
charle [14.2K]2 years ago
6 0
I gotchu

The perimeter is 35. If we were to change the width, which is one of the dimensions of the flower bed, The perimeter will change. This means that perimeter will no longer be 35. So in order to keep the perimeter as it is, if we change one dimension, we must also change the other.

Let's solve for the length, using the formula to see how much the length changes from.

p = 2l + 2w

35 = 2l + 2(15)

35 = 2l + 30

5 = 2l

2.5 = l


We must increase the length from 2.5 feet. This is because decreasing one dimension will decrease the perimeter. But if we increase the other dimension as well, it will restore the perimeter to where is was initially.
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There are 16 pieces of fruit in a basket, including 2 nectarines. What is the probability that a randomly selected piece of frui
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Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!
Akimi4 [234]

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\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill

3 0
3 years ago
At a certain high school, 15 students play stringed instruments and
Aleks [24]

Answer:

<em>0</em> is the probability that a  randomly selected student plays both a stringed and a brass  instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N(A \cup B)' = 5

<u>To find:</u>

The probability that a randomly selected students plays both = ?

<u>Solution:</u>

Total Number of students = N(A)+N(B)+N(A \cup B)' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

N(A\cap B) =0

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0

So, <em>0</em> is the probability that a  randomly selected student plays both a stringed and a brass  instrument.

5 0
2 years ago
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