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3 years ago

Can somebody help me in this question? It would be very appreciated! :)

Mathematics

1 answer:

Fiesta28 [93]3 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

The area (A) of a rectangle is calculated as

A = width × length

let width be w then length is 2w + 3 ( 3 more than twice the width ), then

A = w(2w + 3) ← distribute

= 2w² + 3w

When w = 4 , then

A = 2(4)² + 3(4)

= 2(16) + 12

= 32 + 12

= 44 ft²

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Solve for the given time.

Fittoniya [83]

Answer:

45158

Step-by-step explanation:

Since the colony grows at a rate of 12%, after 2 years there will be $36000\cdot 1.12^2 = 45158.4$ bacteria. Since you probably can't have part of a bacteria, you round to 45158.

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2 years ago

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beks73 [17]

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2 years ago

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There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit

DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = $nCr * p^{r} * q^{n-r}$

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = $15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}$

p(x=3) = 455 * $0.4^{3} * 0.6^{12}$

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337

3 0

2 years ago

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

2 years ago

Rewrite it in factored form: x4 - y2 x2

SCORPION-xisa [38]

Answer:

$(x^2+xy)(x^2-xy)$

Step-by-step explanation:

I'm going to write it so I can see it better:

$x^4-y^2x^2$

You can use the difference of squares formula:

$(x^2+xy)(x^2-xy)$

3 0

2 years ago