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White raven [17]
3 years ago
6

1. What are the solutions (coordinate points) to the system of equations?

Mathematics
1 answer:
Leto [7]3 years ago
3 0

Question 1

The given system of equations is:

y =  {x}^{2}  + 5x + 6 \\ y = 3x + 6

Equate the two equations:

{x}^{2}  + 5x + 6 = 3x + 6

Rewrite in standard form:

{x}^{2}  + 5x - 3x + 6 - 6 = 0

{x}^{2}   + 2x = 0

x(x + 2) = 0

x = 0 \: or \: x =  - 2

When we put x=0, in y=3x +6, we get:

y = 3(0) + 6 = 6

One solution is (0,6)

When we put x=-2, into y=3x+6, we get:

y = 3( - 2) + 6 = 0

Another solution is (-2,0)

The solutions are; (0,6) and (-2,0)

Question 2:

The function is

f(x) =  {x}^{6}  -  {x}^{4}

Let us put x=-x,

f( - x) =  {( - x)}^{6}  -  {( - x)}^{4}

This gives:

f( - x) =  {x}^{6}  -  {x}^{4}

We can observe that:

f(x) = f( - x)

This is the property of an even function.

Question 3:

The given function is

f(x) =  {x}^{2}  + 3x - 2

The average rate of change of f(x) from x=a to x=b is given as:

\frac{f(b) - f(a)}{b - a}

This is the slope of the secant line connecting the two points on f(x)

From x=2 to x=6, the average rate of change

=  \frac{f(6) - f(2)}{6 - 2} \\   =  \frac{ {6}^{2}  + 3 \times 6 -  2 -  {2}^{2}   -  3 \times 2  + 2 }{4} \\   =  \frac{36 + 18 - 4  - 6}{4}  \\  =  \frac{44}{4}   \\ = 11

The average rate of change is 11

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Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

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P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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