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frozen [14]
3 years ago
8

Please help with math.Best answer gets brainiest answer.Please show work.

Mathematics
1 answer:
Andru [333]3 years ago
5 0
Theoretical is when all have the same probability with is 1/6 = .1667 = 16.7% chance

Actual is 12/50 = 24% Which is A!
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Mia wants to borrow £6000 and repay it, with interest, after two years. she sees two offers for loans: offer 1 compound interest
AURORKA [14]

Answer:

she is wrong, offer 2 results in lower interests

Step-by-step explanation:

total amount paid if offer 1 is accepted:

$6,000 x (1 + 3%)² = $6,000 x 1.0609 = $6,365.40

she will pay $365.40 in interests

total amount paid if offer 2 is accepted:

($6,000 x 1.01) x 1.05 = $6,060 x 1.05 = $6,363

she will pay $363 in interests

Compounding interest refers to interest that earns more interest itself, e.g. in the first offer, the $180 of interests charged for the first year will earn $5.40 in extra interests. While offer 2 only charges $60 in interests during the first year which will in turn earn $3 of interests. The difference between both offers is that interest charges in offer 1 earn more interests than the interest in offer 2 = $5.40 - $3 = $2.40

4 0
3 years ago
What is the quotient in simplest form?<br><br> 8 Over 3 divided by two-thirds
Darina [25.2K]

Answer:

4

Step-by-step explanation:

8/3 ÷ 2/3

8/3 * 3/2

4

7 0
3 years ago
Find the GCF for the list.<br> 18y3, 45y²
vlabodo [156]
Since 9 goes into 18 and 45, and both of the integers has a y2, the GCF is 9y2
3 0
4 years ago
22) (2n-q)-56-2,4nt4)
Mumz [18]

Answer:

?

Step-by-step explanation:

I think some of the question is missing.

3 0
3 years ago
For the function​ below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin
Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12

but  

1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}

so the upper sum equals

\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12

When n\rightarrow \infty both \displaystyle\frac{3}{n} and \displaystyle\frac{1}{n^2} tend to zero and the upper sum tends to

\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}

8 0
4 years ago
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