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3 years ago

Not sure how to do this.

Mathematics

1 answer:

Rufina [12.5K]3 years ago

8 0

Answer:

200π

Step-by-step explanation:

Surface area is abbreviated to SA

TOTAL SA of cone = Base area + πrl

Base area = πr² = π x 8² = 64π

Curved SA = πrl = π x 8 x l

to work out l, use pythagoras theorem: √(15)² + (8)² = 17

so, curved SA = π x 8 x 17 = 136π

Total SA of cone = 64π + 136π = 200π

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2 years ago

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What is the answer to this question?

WARRIOR [948]

Answer: Lower right corner

============================================

Explanation:

The information that the temperature rises 3 degrees per hour will tell us the slope here. The slope is rise/run = 3/1 = 3. Each time we move up 3 on the y axis, we move 1 spot to the right.

As you can see in the bottom right hand corner graph, we start off at a temperature of 20 degrees. This is the point (x,y) = (0,20)

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3 years ago

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Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$