All categories

3 years ago

The distance around a circle

Mathematics

2 answers:

sdas [7]3 years ago

5 0

Answer:

The distance around a circle is called perimeter uwu! ^^

brilliants [131]3 years ago

4 0

Answer:

The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference (c). A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d)

Step-by-step explanation:

The formula is simply this: C = πd. In this equation, "C" represents the circumference of the circle, and "d" represents its diameter. That is to say, you can find the circumference of a circle just by multiplying the diameter by pi.

Calculating the circumference of a circle (Pre-Algebra, More)

You might be interested in

How would the average rate of change over years 1 to 5 and years 6 to 10 be affected if the population increased at a rate of 8%

Sever21 [200]

Answer:

(A): As year increases the number of pikas reduces.

(B): As year increases the number of pikas increases as opposed to when the rate reduces.

Step-by-step explanation:

See comment for complete question

Given

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

(A) WHEN THE RATE DECREASES

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1-8\%)^x$

Convert % to decimal

$f(x) = 144(1-0.08)^x$

$f(x) = 144(0.92)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(0.92)^x$

$f(1) = 144*0.92^1 =144*0.92=132.48$

$f(5) = 144*0.92^5 =144*0.66=95.04$

$Rate = \frac{95.04 - 132.48 }{4}$

$Rate = \frac{-37.44}{4}$

$Rate = -9.36$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(0.92)^x$

$f(6) = 144*0.92^6 =144*0.61=87.84$

$f(10) = 144*0.92^{10} =144*0.43=61.92$

$Rate = \frac{61.92-87.84}{4}$

$Rate = \frac{-25.92}{4}$

$Rate = -6.48$

So, we have:

$Rate = -9.36$ for year 1 to 5

This means that the number of pikas reduces by 9.36 yearly

$Rate = -6.48$ for year 6 to 10

This means that the number of pikas reduces by 6.48 yearly

So, we can say that, as year increases the number of pikas reduces.

(B) WHEN THE RATE INCREASES

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1+8\%)^x$

Convert % to decimal

$f(x) = 144(1+0.08)^x$

$f(x) = 144(1.08)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(1.08)^x$

$f(1) = 144(1.08)^1 = 144*1.08= 155.52$

$f(5) = 144(1.08)^5 = 144*1.47= 211.68$

$Rate = \frac{211.68 - 155.52}{4}$

$Rate = \frac{56.16}{4}$

$Rate = 14.04$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(1.08)^x$

$f(6) = 144(1.08)^6 = 228.52$

$f(10) = 144(1.08)^{10} = 310.89$

$Rate = \frac{310.89-228.52}{4}$

$Rate = \frac{82.37}{4}$

$Rate = 20.59$

So, we have:

$Rate = 14.04$ for year 1 to 5

This means that the number of pikas increases by 14.04 yearly

$Rate = 20.59$ for year 6 to 10

This means that the number of pikas increases by 20.59 yearly

So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.

5 0

3 years ago

Two teaching methods and their effects on science test scores are being reviewed. A random sample of 18 students, taught in trad

Leona [35]

Answer:

the value of the t test statistic is - 3.419

Step-by-step explanation:

Given that;

n₁ = 18, u₁ = 78.3, s₁ = 6.4

n₂ = 11, u₂ = 84.3, s₂ = 5.3

α = 0.1

Now The hypothesis are;

H₀ : u₁ = u₂

H₁ : u₁ < u₂

To compute the value of the t test statistic;

t = [(x₁ - x₂) / s × √(1/n₁ + 1/n₂)]

where

s = √ [ ((n₁-1) × s₁² + (n₂ - 1 ) × s₂²) / ( n₁ + n₂ - 2)]

s = √ [ ((18-1) × 6.4² + (11 - 1 ) × 5.3²) / ( 18 + 11 - 2)]

s = √ [ (7 × 40.96 + 10 × 28.09 ) / 27 ]

s = √ [ (286.72 + 280.9) / 27 ]

s = √(567.62/27)

s = √21.0229

s = 4.585

Now t test statistics t = [(x₁ - x₂) / s × √(1/n₁ + 1/n₂)]

t = [(78.3 - 84.3) / 4.585 × √(1/18 + 1/11)]

t = -6 / (4.585 × 0.3827)

t = - 6 / 1.7546795

t = - 3.419

Therefore the value of the t test statistic is - 3.419

as as level of significance α = 0.1

df = 18+11-2 = 27

∴ T(csal) = t(0.1, 27) = -1.313

That is

t(statistics) < t(cal)

{ - 3.419 < -1.313 }

5 0

3 years ago

14 + w is the same as which expression?

Dvinal [7]

Answer:

I think it's w + 14? :3

AND I THOUGHT I SAW A KUNIMI PFP-

Step-by-step explanation:

8 0

3 years ago

Simple question...but yet i'm unable to answer.... help please?....

kati45 [8]

T=50 ithinkso right yes

3 0

4 years ago

Paloma has a flower stand at a farmer's market. On Tuesday, she sold 8 marigold flats and 10 daisy flats. An hour before closing

Nostrana [21]

Answer:

She made $117.25

Step-by-step explanation:

Consider the provided information.

The price of marigold(M) is $5.75.

The price of daisy(D) is $6.25.

The price of roses(R) is $2.50.

The price of Lilies(L) is $3.75

She sold 8 marigold flats and 10 daisy flats. An hour before closing, she sold 5 bunches of roses at half price and 2 bunches of lilies for 1/3 of the price

Total money Paloma made = $8M+10D+5\times \frac{R}{2}+2\times \frac{L}{3}$

Now substitute the respective value

Total money Paloma made = $8(5.75)+10(6.25)+5\times \frac{2.50}{2}+2\times \frac{3.75}{3}$

Total money Paloma made = $46+62.5+6.25+2.5$

Total money Paloma made = $117.25$

Hence, she made $117.25

3 0

3 years ago