Answer:
49.7%
Step-by-step explanation:
A cdircle is located within a square.
<u>Area of the circle</u>
Area = 
, where r = 4 units.
Area Circle = 50.3 units^2
<u>Area of the square</u>
Area = l*w or l^2 for a square, since l = w
Area = (10 units)^2
Area = 100 units^2
<u>Area in the square but outside the circle</u>
This is the difference [Square minus Circle Areas]
Square minus Circle Areas = 100 - 50.3 or <u>49.7 units^2</u>
<u>Probability</u>
The probability of picking a point in the square that is not in the circle is the ration of the two areas: <u>[Outside Circle/Square]x100%</u>
<u></u>
<u>(</u>49.7 units^2)/(100 units^2)x100% = 49.7%
<u></u>
Answer:
if everything cost the same thing you have to do 3x5 to get the answer so
3x5=$15 + the $3 kids meal sot the answer is 18. and eand up with 10$
Step-by-step explanation:
The expected value for the person buying the insurance is -25.
<h3>How the expected value is calculated?</h3>
The expected value is the average gain or loss of an event if the event is repeated a number of times.
Expected value = ∑xP(x)
<h3>Calculation:</h3>
It is given that,
The probability of a 47-year-old woman passing away during the coming year is 0.179% = 0.00179
The death benefit = $100,000 - $204 = $99,796
The loss from living = -204
Then the expected value = 99796(0.00179) + (-204)(0.99821) = -25
Therefore, the expected value for the person buying the insurance is -25.
Learn more about the expected value here:
brainly.com/question/10675141
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The sequence can be represented by 
, where 
is a whole number(
and is an integer). By this, we should be able to plug in some whole number x and be able to get these values if they are in the sequence. To test this, let's set these numbers equal to and confirm that there is a whole number x which equals the values selected.
When trying 160 with our equation, we find that x doesn't equal a whole number, thus our answer is C, or 160. To confirm our answer, we can test 170 and 210 and confirm that they both give us a value for x which is a whole number.
