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3 years ago

Please help with this ?!?

Mathematics

1 answer:

Tatiana [17]3 years ago

4 0

Radius = 1/2 Diameter
Diameter = 12
Radius = 6

Area of circle = (pi) r ^2
=pi 6^2
=36pi

Area = 113.1 cm^2

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Sean is taking drivers ed. And the instructor asked him to stay within 2 miles of the posted speed limits. The current speed lim

Aleonysh [2.5K]

Answer:

{x| 43 ≤ x ≤ 47} where x = Sean's speed (in mph)

Step-by-step explanation:

Sean is taking driver's ed. The instructor asked him to stay within 2 miles of the posted speed limits. The current speed limit is 45 mph.

So, the maximum speed which Sean can reach is, (45 + 2) mph

=47 mph

and, the minimum speed which Sean can reach is, (45 - 2) mph

=43 mph

so, {x| 43 ≤ x ≤ 47} where x = Sean's speed (in mph)

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3 years ago

916 round it to the nearest hundreds

dlinn [17]

Answer:

900

Step-by-step explanation:

9<u>1</u>6

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3 years ago

Read 2 more answers

Evaluate the following, if a=7, b=3, c=2

Zinaida [17]

Answer:

Step-by-step explanation:

a^2 - 5bc - 1

a = 7

b = 3

c = 2

7^2 - 5*3*2 - 1

49 - 30 - 1

4 0

2 years ago

If f(x)=3x + 6 then what is the inverse of f(x)

kakasveta [241]

Inverse all of the signs it will be (x-6)/3

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3 years ago

Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How

HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

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.

Parameter of 5.2 per square yard:

This means that $\mu = 5.2r$ , in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

$P(X \geq 1) = 1 - P(X = 0) = 0.99$

Thus:

$P(X = 0) = 1 - 0.99 = 0.01$

We have that:

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

$P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}$

Then

$e^{-5.2r} = 0.01$

$\ln{e^{-5.2r}} = \ln{0.01}$

$-5.2r = \ln{0.01}$

$r = -\frac{\ln{0.01}}{5.2}$

$r = 0.89$

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0

3 years ago