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3 years ago

How much does the CEO earns?

Mathematics

2 answers:

melomori [17]3 years ago

8 0

p = number of items sold (note: this is not the profit)

1 item sells for $855, so p items sell for 855p dollars

Subtract off the cost of 6780 and we have the expression 855p-6780 which is the profit for that given month.

Now plug in p = 250 because 250 items were sold in that given month

855*p - 6780 = 855*250 - 6780 = 206,970

The company earns $206,970 in profit for that month. Apply 15% to this value

15% of 206,970 = (15/100)*206,970 = 0.15*206,970 = 31,045.50

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Answer: $31,045.50 which is choice C

tresset_1 [31]3 years ago

5 0

Answer:
The CEO earns a total of $31,045.50

Explanation:

855 * 250 = $213,750

213,750 - 6,780 = 206,970

206,970 * 15% = $31,045.50

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Mary invests £12000 in a savings account. The account pays 1. 5% compound interest per year. Work out the value of her investmen

mestny [16]

Answer:

The value of Mary's investment after two years = £12362.7

Step-by-step explanation:P = Principal / initial amountR = rate of interest per cent per yearT = number of yearsA = final amount at the end of T yearsThen:A = P*(1 + R/100)^2In our example:P = £12000R = 1.5 per cent per yearT = 2 yearsThus:A = 12000*(1 + 1.5/100)^2 = 12000*(1 + 0.015)^2 = 12000*(1.015)^2 = 12000*(1.030225) = 12362.7Value of investment after two years = £12362.7

4 0

2 years ago

a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter

andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an|, converges if $0< b_{n+1} \leq b_n$ for all n, and $\lim_{n \to \infty} b_n =$ 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

Part a

An Alternating series is an infinite series given on these three possible general forms given by:

$\sum_{n=0}^{\infty} (-1)^{n} b_n$

$\sum_{n=0}^{\infty} (-1)^{n+1} b_n$

$\sum_{n=0}^{\infty} (-1)^{n-1} b_n$

For all $a_n >0, \forall n$

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

Part b

An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an| converges if $0< b_{n+1} \leq b_n$ for all n and $\lim_{n \to \infty} b_n$ =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

Theorem (Alternating series test)

If a sequence of positive terms {bn} is monotonically decreasing and

$\lim_{n \to \infty} b_n = 0$ , then the alternating series $\sum (-1)^{n-1} b_n$ converges if:

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

then $\sum_{n=1}^{\infty}(-1)^{n-1} b_n$ converges

Proof

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

$|s_n -s| < \epsilon$ for all n, and that implies that the series converges to a value, s.

And this complete the proof.

Part c

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where $s_n$ represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most $b_{n+1}$ by the following theorem:

Theorem (Alternating series sum estimation)

If $\sum (-1)^{n-1} b_n$ is the sum of an alternating series that satisfies

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

Then then $\mid s - s_n \mid \leq b_{n+1}$

Proof

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1.

$\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}$