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2 years ago

Hi pls help me with this problem

Mathematics

1 answer:

ipn [44]2 years ago

3 0

Answer:

Why is base pairing essential to the process of transcription and translation

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When the effective interest rate is 9% per annum, what is the present value of a series of 50 annual payments that start at $100

ser-zykov [4K]

Answer:

$1,109.62

Step-by-step explanation:

Let's first compute the future value FV.

In order to see the rule of formation, let's see the value (in $) for the first few years

End of year 0

1,000

End of year 1(capital + interest + new deposit)

1,000*(1.09)+10

End of year 2 (capital + interest + new deposit)

(1,000*(1.09)+10)*1.09 +10 =

$\bf 1,000*(1.09)^2+10(1+1.09)$

End of year 3 (capital + interest + new deposit)

$\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)$

and we can see that at the end of year 50, the future value is

$\bf FV=1,000*(1.09)^{50}+10(1+1.09+(1.09)^2+...+(1.09)^{49}$

The sum

$\bf 1+1.09+(1.09)^2+...+(1.09)^{49}$

is the sum of a geometric sequence with common ratio 1.09 and is equal to

$\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356$

and the future value is then

$\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564$

The present value PV is

$\bf PV=\frac{FV}{(1.09)^{50}}=\frac{82508.35564}{74.35572}=1,109.616829\approx \$1,109.62$

rounded to the nearest hundredth.

5 0

3 years ago

Jadyn prepares a bag of candy before a road trip. The bag contains individual pieces of Skittles, M&Ms, Life Savers and Sour

kipiarov [429]

Given condition is- The bag contains individual pieces of Skittles, M&Ms, Life Savers and Sour Patch Kids.

Let the total number of candies in the bag be = x

Then, Skittles are $\frac{1x}{2}$

Candies left = $x-\frac{x}{2}$ = $\frac{x}{2}$

M&Ms are = $\frac{1x}{8}$

Now remaining candies are= $\frac{3x}{8}$

Now Life savers are = $\frac{9x}{32}$

Now adding all these 3 candies

$\frac{x}{2}+ \frac{x}{8}+\frac{9x}{32}$

= $\frac{29x}{32}$

Now the remaining candies sour patch will be =

$x-\frac{29x}{32} =\frac{3x}{32}$

Hence, sour patch kids are $\frac{3}{32}$

3 0

3 years ago

Line M passes through the points (5,1) and (8,6) while line n passes through the point (-4,3) and (-1,8) . Which statement accur

kirill [66]

M=(6-1)/(8-5)=5/3

y=5x/3+b, using (5,1)

1=25/3+b

b=-22/3

y=(5x-22)/3

...

m=(8-3)/(-1--4)=5/3...

y=5x/3+b, using (-1,8)

8=-5/3+b

b=29/3

y=(5x+29)/3

Since they have the same slope and difference y-intercepts they are parallel lines and will never intersect.

4 0

3 years ago

Read 2 more answers

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

3 years ago

Marla has $95.50 in her savings account. She deposits $5 every week. Her mother also deposits $20 into the account every time Ma

vladimir2022 [97]

Part A: Coefficients: 20,5.
Variables: c,w
Constants:95.50

6 0

3 years ago