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2 years ago

a circle has a diameter of 10cm what is the best proximation of this area use 3.14 to approximate for pi

2 answers:

5 0

The meaning of the area is the measurement of a surface in this case a circle, which can be find by using the formula A=πr^2, where r is equal to the radius of the circle.

On this exercise is given that a circle has a diameter of 10cm and it is asked to find its area, as previous said the only way to find the area of a circle is to use the formula A=πr^2, but on this case the radius is not given. As result you have to divide the measurement of the diameter by two, and the number found represents the radius.

10cm/2= 5cm The radius of the circle is 5cm.

Now that the radius is known you can substitute the values in the formula to find the area of the circle.

A=πr^2
A=(3.14)(5)^2
A=(3.14)(25)
A=78.5 cm^2

The area of the circle is 78.5 square centimeters.

hammer [34]2 years ago

4 0

Hello!

The answer would be $75.8^{cm2}$

Hope this helped!

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A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic

aliya0001 [1]

Answer:

P-value = 0.4846

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.73

Sample mean, $\bar{x}$ = 4.35

Sample size, n = 51

Alpha, α = 0.05

Sample standard deviation, s = 3.88

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73$

We use Two-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699$

Now, $z_{critical} \text{ at 0.05 level of significance } = \pm 1.96$

We can calculate the p-value with the help of standard normal table.

P-value = 0.4846

Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.

We conclude that this college has same drinking habit as the college students in general.

7 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.