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mina [271]
3 years ago
13

Question 28

Mathematics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

Given Rate of interest is r=8%=0.08

Principal Amount is A=5,000

Time is t years

Interest is compounded yearly twice ⟹n=2

Amount =P(1+

n

r

)

nt

=5000×(1+

2

0.08

)

2t

=5,408

(1.0816)

t

=

5000

5408

⟹t=1

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7.6 liters

Step-by-step explanation:

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3 years ago
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In three separate stock trades Dale los $560, gained $850, and lost $280. What was the net result of the three trades?
romanna [79]

Answer:

$10 profit

Step-by-step explanation:

560+280=840

850-840=10

5 0
2 years ago
The sum of the first 15 terms of a geometric sequence with 7 as the first term and a common ratio of -3 is __________.
Andre45 [30]

Step-by-step explanation:

the formula for the sum of the first n terms of a geometric sequence is

Sn = s1(1 - r^n)/(1 - r)

with r being the common ratio and s1 is the first term.

so,

S15 = 7×(1 - (-3)¹⁵)/(1 - -3) = 7×(1 - -14,348,907)/4 =

= 7×14,348,908/4 = 7×3,587,227 = 25,110,589

4 0
2 years ago
Please help me with this thank you
Varvara68 [4.7K]
Too much sand.

Volume is the amount of space something takes up. The formula for volume is v=length x width x height.

The volume of the sandbox is 36 ft^3, because 5x6x1.2 = 36. This means the box takes up 36 cubic feet of space.

If the customer bought 40 cubic feet of sand, than they bought to much because 40 > 36. The amount of sand she got is greater than the volume of the box, so it won’t all fit.
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2 years ago
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Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
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