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3 years ago

F(x)=2x^2-x-10 part a: what are the X intercepts of the graph of f(x)? Show your work

1 answer:

4 0

A. x intercepts are where the graph hits the x axis or where f(x)=0
0=2x^2-x-10
solve
hmm
we can use the ac method
multiply 2 and -10 to get -20
what 2 numbers muliply to get -20 and add to get -1 (the coefint of the x term)
-5 and 4
split the midd into that
0=2x^2+4x-5x-10
group
0=(2x^2+4x)+(-5x-10)
factor
0=2x(x+2)-5(x+2)
undistribute
0=(x+2)(2x-5)
set each to 0
0=x+2
0=-2

0=2x-5
5=2x
5/2=x

x intercepts are at x=-2 and 5/2 or the points (-2,0) and (5/2,0)

B. ok, so for f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
if a<0 then the parabola opens down and the vertex is a max

f(x)=2x^2-x-10
2>0
opoens up
vertex is minimum

ok, the vertex

the x value of the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a)) so
given
f(x)=2x^2-x-10
a=2
b=-1
-b/2a=-(-1)/(2*2)=1/4
f(1/4)=2(1/4)^2-(1/4)-10
f(1/4)=2(1/16)-1/4-10
f(1/4)=1/8-1/4-10
f(1/4)=1/8-2/8-80/8
f(1/4)=-81/8
so the vertex is (1/4,-81/8) or (0.25,-10.125)

C. graph the x intercepts and the vertex
the vertex is min and the graph goes through the x intercepts

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Answer:

B and D

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Use the graph to estimate the solution to the system of equations shown.

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<h3>Answer: D. x= -5 1/2, y= -3 1/4</h3>

==============================================================

Explanation:

The solution occurs where the two lines intersect. From this intersection point, draw a vertical line upward until you reach the x axis. We arrive somewhere between -5 and -6. A good estimate is to say that we're at the midpoint, so x = -5 & 1/2 is one possible estimate

Since only choice D has this x value, this must mean choice D is the answer. The y coordinate of y = -3 & 1/4 seems to fit as well since we can draw a horizontal line from the point of intersection to the y axis, and arrive somewhere just a little below -3.

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3 years ago

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that comput

Andrej [43]

Answer:

a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

Step-by-step explanation:

The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).

It is given by the following formula:

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success.

In this problem, a success is being concerned that employers are monitoring phone calls.

53% of adults are concerned that employers are monitoring phone calls, so $p = 0.53$

(a) Out of four adults, none is concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 0 successes, so x = 0.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,0}.(0.53)^{0}.(0.47)^{4}$

$P = 0.0488$

There is a 4.88% probability that none is concerned that employers are monitoring phone calls.

(b) Out of four adults, all are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 4.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,0}.(0.53)^{4}.(0.47)^{0}$

$P = 0.0789$

There is a 7.89% probability that all are concerned that employers are monitoring phone calls.

(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.

Four adults, so $n = 4$ .

Is the probability of 4 successes, so x = 2.

$P = C_{n,x}.p^{n}.(1-p)^{n-x}$

$P = C_{4,2}.(0.53)^{2}.(0.47)^{2}$

$P = 0.3723$

There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.

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