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3 years ago

Which of the following is a solution to the inequality 7-1/5x>0?

Mathematics

1 answer:

beks73 [17]3 years ago

4 0

$7-\dfrac{1}{5}x>0\qquad\text{subtract 7 from both sides}\\\\-\dfrac{1}{5}x>-7\qquad\text{change the signs}\\\\\dfrac{1}{5}x$

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Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

Salsk061 [2.6K]

Answer: In June 2097

Step-by-step explanation:

According to the model, to find how many years t should take for $P(t)=21100$ we must solve the equation $0.9t^2+6t+14000=21100$ . Substracting 21100 from both sides, this equation is equivalent to $0.9t^2+6t-7100=0$ .

Using the quadratic formula, the solutions are $t_1= \frac{-6-\sqrt{6^2 -4*0.9*(-7100)}}{2*0.9}=-92.21$ and $t_2=\frac{-6+\sqrt{6^2 -4*0.9*(-7100)}}{2*0.9}=85.54$ . The solution $t_1=-92.21$ can be neglected as the time t is a nonnegative number, therefore $t=t_2=85.54$ .

The value of t is approximately 85 and a half years and the initial time of this model is the January 1, 2012. Adding 85 years to the initial time gives the date January 2097, and finally adding the remaining half year (six months) we conclude that the date is June 2097.

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3 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

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