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2 years ago

Area of composite figures worksheet. How do you solve it. Step-By-Step

Mathematics

1 answer:

Grace [21]2 years ago

5 0

Answer:

The area of the combined figure is 49.8 cm

Step-by-step explanation:

We first see that there is a semicircle, which represents half of a full circle. We must find the area of such an entire circle, so we have to divide its entire area by two since only half of such an area is represented:

The area of a circle is given by:

In this case, the radius is given by half the base of the rectangle.

So much so that the radius of that radius is 2.5cm, we now consider the pi to be a constant measure of 3.14

Remember that we are only representing half of the circle, therefore, we must divide the area by two:

<h3>Ac/2 = 19.625 / 2 = 9.8 <em>(</em><em> </em><em>aproximate</em><em> </em><em>)</em></h3>

Now we proceed to find the area of the rectangle, whose height is 8 cm, and base equal to 5 cm.

Since the area of a rectangle is given by multiplying the base times the height, we have:

In this case the figure is complete, there is no need to divide.

To find the full area, we must add the area of the half circle more the area of the newly found rectangle:

<h3>Overall area = 9.8 cm + 40 cm = 49.8 cm</h3>

By so:

<h3>The area of the combined figure is 49.8 cm</h3>

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How many terms does the expression q÷2 +g-2k

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Answer:

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3 years ago

The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th

Nikitich [7]

Answer:

a) $\mu_{\bar X} =13.35$

e.none of the above

b) $\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02$

a.0.0200

c) $z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5$

$P(\bar X< 13.32)= P(z$

a.0.0668

d) $z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5$

$z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5$

$P(13.30$

c.0.6853

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(13.35,0.12)$