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Anni [7]
3 years ago
14

Can someone help me figure out the answer?

Mathematics
1 answer:
olga_2 [115]3 years ago
8 0

Answer:

y = -2x +15

Step-by-step explanation:

The point-slope form of the equation for a line through (h, k) with slope m is ...

... y - k = m(x - h)

For your point (h, k) = (5, 5) and slope m = -2, the equation in point-slope form is ...

... y - 5 = -2(x - 5)

Simplifying, we get

... y - 5 = -2x +10

Adding 5 puts the equation into slope-intercept form, as you want.

... y = -2x +15

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So here, you just put in the value inside the parenthesis of the function into the actual function itself.

Here, for h(f(2)), the first one to put in is the one in f, which is 2.

Since f(x) = 3*2^x, and since x here is equal to 2, (because h(f(2)), we can just put in the 2 in place of the x.

3 * 2 ^ 2

And then you get:

3 * 4 

Or 12.

So now you have

h(12)

Now do the same thing.

2 * 12 - 7

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17
8 0
3 years ago
Dan is watching the birds in his backyard. Of the birds he watches, 9 of them, or 45%, are sparrows. How may birds are in his ba
andrezito [222]

Answer:

5

Step-by-step explanation:45 divided by 9 is 5

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5 points is not enough

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Of 60 randomly chosen students from a school surveyed,16 chose aquarium as their favorite fiel trip.There are 720 students in th
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Answer:

192 students

Step-by-step explanation:

60 randomly students surveyed

16 chose aquarium as their favorite field trip.

720 students in total.

\frac{16}{60} = \frac{?}{720} =

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5 0
3 years ago
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul
Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week, \mu = \frac{3}{20} = 0.15

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

P(X \leq 1) = P(X = 0) + P(X = 1)

Then

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607

P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

6 0
2 years ago
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