Sharyce put some money in a savings account that paid 2.7% simple interest. After a year, the account was worth $7,890.44. What
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Let
r------> the radius of the circular tabletop
we know that
so
The area of a circle is equal to
where
r is the radius of the circle
<u>Part a)</u> What is the smallest possible area of the tabletop that will fit on Timothy’s table base?
we know that
the smallest possible area of the tabletop is for
Substitute the value of r in the formula
Round to the nearest whole square inch
so
therefore
<u>the answer part a) is </u>
the smallest possible area of the tabletop is
<u>Part b)</u> What is the largest possible area of the tabletop that will fit on Timothy’s table base?
we know that
the largest possible area of the tabletop is for
Substitute the value of r in the formula
Round to the nearest whole square inch
so
therefore
<u>the answer part b) is </u>
the largest possible area of the tabletop is
Answer:
And we can find this probability with this difference and using the normal standard table or excel:
And the result is illustrated in the figure attached.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference and using the normal standard table or excel:
And the result is illustrated in the figure attached.
Solve equation [1] for the variable u
[1] u = v
// Plug this in for variable u in equation [2]
[2] 8•(v ) - 4v = -32
[2] 4v = -32
// Solve equation [2] for the variable v
[2] 4v = - 32
[2] v = - 8
// By now we know this much :
u = v
v = -8
// Use the v value to solve for u
u = (-8) = -8
Solution : {u , v } = { -8,-8}
[1] u = v
// Plug this in for variable u in equation [2]
[2] 8•(v ) - 4v = -32
[2] 4v = -32
// Solve equation [2] for the variable v
[2] 4v = - 32
[2] v = - 8
// By now we know this much :
u = v
v = -8
// Use the v value to solve for u
u = (-8) = -8
Solution : {u , v } = { -8,-8}