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Reptile [31]
3 years ago
7

The area of the circle is 144π m2 What is the diameter of the circle?

Mathematics
1 answer:
Over [174]3 years ago
5 0

Answer:

Step-by-step explanation:

Remember that

Area of Circle = \pi r^2

\pi  = pi \\r = radius

They told us that for this circle, the area is 144\pi  m^2

If you put it back into the formula for area of a circle we get:

Area of Circle = \pi r^2

144\pi = \pi r^2

Divide both sides by \pi.

144 = r^2

Now we want to solve for r (r = radius).

You need to square root each side

\sqrt{144} = \sqrt{r^2}

12 = r

So now we know the radius, r = 12 metres

Also remember that

Diameter = 2r

So,

Diameter = 2 times 12 = 24 metres

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PLEASE HELP ME OUT? it's an assignment due in an hour!
solmaris [256]

Answer:

f(g(5)) = 11

Step-by-step explanation:

  • f(x) = x - 2
  • g(x) = 3x - 2

The composite function f(g(5)) is found by plugging g(5) for x into f(x).

Let's find g(5) first.

  • g(5) = 3(5) - 2
  • g(5) = 15 - 2
  • g(5) = 13

Next, plug this value for x into f(x).

  • f(13) = (13) - 2
  • f(13) = 11

∴ <u>f(g(5)) = 11</u>.

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2 years ago
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Which number is 247, 039 ruonded to the nearest thousand?
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247,039 rounded to the nearest thousand becomes

247,000

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
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In this questions, there are a few functions with more/less than and equal to symbol. 
1.  Answer: f(x)= 3
f(x)= 2<x ≤3
The x is [more than 2] and [less than 3] or [equal to 3]. The number that will fulfill the condition would be only 3 since 3 is [more than 2] and [equal to 3].

2. f(x)= 1
f(x)= 0<x ≤1
The x is [more than 0] and [less than 1] or [equal to 1]. The number that will fulfill the condition would be only 1 since 1 is [more than 0] and [equal to 1].


3. f(x)= x
f(x)= 1≤x ≤2
The x is [more than 1] or [equal to 1] and [less than 2] or [equal to 2]. The number that will fulfill the condition would be
1, since 1 is [equal to 1] and [less than 2] 
2, since 2 is [more than 1] and [equal to 2] 
The graph that follow the value would be: f(x)= x

4. Answer: f(x)= 4
f(x)= 3<x ≤4
The x is [more than 3] and [less than 4] or [equal to 4]. The number that will fulfill the condition would be only 4 since 4 is [more than 3] and [equal to 4].
7 0
3 years ago
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a seamstress has a piece of cloth that is 3 yards long. she cuts it into shorter lengths of 16inches each. how many shorter piec
NeTakaya

Answer

Find out the how many shorter pieces can a seamstress cut.

To prove

Let us assume that the number of shorter pieces  cut be x.

As given

A seamstress has a piece of cloth that is 3 yards long.

she cuts it into shorter lengths of 16 inches each.

As

1 yards = 36 inches

Now convert 3 yards into inches.

= 3 × 36

= 108 yards

Than the equation becomes

16x = 108

x = \frac{108}{16}

x = 6.75

Therefore the number of shorter pieces be 6 .




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3 years ago
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Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
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