B. what is a real-world example or use of surface area?

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

5 0

Answer:

Surface area : It is the total area of a surface

The surface area is used in many places.

1) If we want to paint a wall, we have to find the total surface area of the wall.

2). If we want to place tiles in our floor we want to find surface area.

3)If we want to roof our houses we have to find the surface area.

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Will award brainliest to correct answer!

Artyom0805 [142]

Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

$\angle \: ABP = \frac{180 - 80}{2} = 50 \degree$

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)

6 0

2 years ago

The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen

faltersainse [42]

Answer:

$a=150.7 -0.385*10.2=146.773$

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.

Step-by-step explanation:

1) 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".

2) Solution to the problem

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

$X \sim N(150.7,10.2)$

Where $\mu=150.7$ and $\sigma=10.2$

We want to find a value a, such that we satisfy this condition:

$P(X>a)=0.65$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

If we use condition (b) from previous we have this:

$P(X$