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Ad libitum [116K]
3 years ago
10

A circle has an area of 225pift2 find the radius of the circle and find the diameter and circumference of the circle

Mathematics
1 answer:
RideAnS [48]3 years ago
6 0

Answer:

  • radius: 15 ft
  • diameter: 30 ft
  • circumference: 30π ft

Step-by-step explanation:

The area is given in terms of the radius as ...

  A = πr^2

so we can find the radius as ...

  r = √(A/π)

For your circle, the radius is ...

  r = √(225π ft^2/π) = 15 ft

The diameter is twice the radius, so is ...

  d = 2r = 2·15 ft = 30 ft

And the circumference is π times the diameter:

  C = πd = π·30 ft = 30π ft

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A rectangular parking lot must have a perimeter of 440 feet and an area of at least 8000 square feet. Describe the possible leng
tangare [24]

Answer:

The possible parking lengths are 45.96 feet and 174.031 feet

Step-by-step explanation:

Let x be the length of rectangular plot and y be the breadth of rectangular plot

A rectangular parking lot must have a perimeter of 440 feet

Perimeter of rectangular plot =2(l+b)=2(x+y)=440

2(x+y)=440

x+y=220

y=220-x

We are also given that an area of at least 8000 square feet.

So, xy \leq 8000

So,x(220-x) \leq 8000

220x-x^2 \leq 8000

So,220x-x^2 = 8000\\-x^2+220x-8000=0

General quadratic equation : ax^2+bx+c=0

Formula : x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

x=\frac{-220 \pm \sqrt{220^2-4(-1)(-8000)}}{2(-1)}\\x=\frac{-220 + \sqrt{220^2-4(-1)(-8000)}}{2(-1)} , \frac{-220 - \sqrt{220^2-4(-1)(-8000)}}{2(-1)}\\x=45.96,174.031

So, The possible parking lengths are 45.96 feet and 174.031 feet

3 0
3 years ago
Is (-1,2) a solution of y=-2x-1
Klio2033 [76]
A solution of a function is where the function hits the x-axis, so if y is any number that is not equal to 0, there is no possible way that point can be a solution. 

Hope this helps!
7 0
3 years ago
Read 2 more answers
A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
Explain the difference between each pair of concepts.
Fynjy0 [20]

Answer:

a. B. Frequency is the number of times a particular distinct value occurs. Relative frequency is the ratio of the frequency of a value to the total number of observations.

b. A. A relative frequency is the same as a percentage expressed as a decimal.

Step-by-step explanation:

Frequency is the number of repetitions of a particular observation.

Relative Frequency is the ratio of the frequency of particular observation to the sum of frequencies of all the observations.

The percentage is the proportion of the whole.

Thus, In question (a)

Option B is the only correct answer.

and in question (b)

Option A is the only correct answer.

7 0
3 years ago
Help me idk what these are
ryzh [129]
1. is 23

2. is 29

3. is A

4. is A
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3 years ago
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