All categories

3 years ago

5 A dangerous criminal has escaped from

Mathematics

1 answer:

zepelin [54]3 years ago

4 0

Answer:

314 mile²

Step-by-step explanation:

area = πr² = 3.14 x 10² = 314 mile²

You might be interested in

Help me with this one problem please :( Find the following arc measures.

r-ruslan [8.4K]

Answer:

Arc KL $=23$

Arc OM $=113$

Step-by-step explanation:

We know that Angle NPM is equal to 90 degrees.

Therefore, Angle KPM is also equal to 90 degrees.

We are told that Arc LM is equal to 67.

Thus, the measure for Arc KL is equal to $90-67=23$

We also know that Angle OPN is equal to Angle KPL.

Therefore, the measure of Arc ON $=23$

Using that information, we know that Arc OM is equal to $90+23=113$

Hope I helped!

3 0

2 years ago

Statements showing a relationship between numbers that are not necessarily equal using the symbols >, <, ≥, ≤, or ≠ are ca

likoan [24]

3.) Inequalities.
Hope it helped.

7 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago

Really never learned this

kykrilka [37]

The perimeter of a two-dimensional figure is always the sum of its sides.

The exercise mat has a perimeter (P) of 36 feet. ...

Let length be L, and width be W.

P = 36

L = 2w

P=2(l+w)=2·(12+6)=36

L = 12 feet

W = 6 feet

8 0

2 years ago

Will give brainlest answer please help

Tju [1.3M]

Answer:

-1

Step-by-step explanation:

You can do this two ways, the easiest way is to look at the graph.

What is g(4) means: When the x value equals 4, what does y equal?

Look for the coordinate with the x-value of 4 (it's asking for g(4). Now if you look at the y-value of the coordinate, it should tell you what g(4) is. So since the coordinate is (4,-1), g(4) is -1.

Another way to do it is to plug in 4 for x in the equation. So...

g(4) = -[f(4) - 3)]

g(4) = -[4 - 3]

g(4) = -[1]

g(4) = -1

7 0

2 years ago