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Nezavi [6.7K]
3 years ago
13

Find the circumference and area of each circle. Include the formula thank you

Mathematics
2 answers:
yaroslaw [1]3 years ago
8 0
Circumference=16 pi
Area=64 pi
Below is my methods

Hitman42 [59]3 years ago
8 0

Answer:

Circumference: 50.27 cm

Area: 201.06 cm²

Step-by-step explanation:

Hi there!

<u>1) Determine the circumference</u>

C=\pi d where d is the diameter of the circle

Plug in the diameter (16 cm)

C=16\pi\\C=50.27

Therefore, the circumference is 50.27 cm when rounded to the nearest hundredth.

<u>2) Determine the area</u>

A=\pi r^2 where r is the radius

Plug in the radius (16/2=8)

A=\pi (8)^2\\A=64\pi\\A=201.06

Therefore, the area of the circle is 201.06 cm² when rounded to the nearest hundredth.

I hope this helps!

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Can someone PLEASE help me!! I'm doing a test and I'm really stuck of this question! Each of the small squares in the figure bel
Arturiano [62]

Answer:

40 meters²

Step-by-step explanation:

Formula for the area of a triangle is \frac{1}{2} *b*h

\frac{1}{2} *10*8=\\5*8=\\40

The area of the triangle is 40 m²

7 0
3 years ago
Help, can’t figure it out, need help fast
olga55 [171]
B to the second means b * b.
The short hand for b*b is b^2 on this editor or b^{2}
c to the fourth means c * c * c *c
the short hand for c*c*c*c is c^4 on this editor or  c^{4}

The answer you want is b^2/c^4 in the answer box. 
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5 0
3 years ago
Assume that a student is chosen at random from a class. Determine whether
antiseptic1488 [7]

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Mutually exclusive

Step-by-step explanation:

8 0
3 years ago
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natta225 [31]

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5 0
3 years ago
A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
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