All categories

3 years ago

If ƒ={(5, 1),(6, 2),(7, 3),(8, 1),(9, 7)}, then the range of ƒ is

2 answers:

7 0

1,2,3,7
I believe the range is just depending on your y values,and since an equation is not given your range should just be the y values of that set

son4ous [18]3 years ago

3 0

I think this should help :-)

1. The points that are in given in the (x, y) form aka (x-coordinate, y-coordinate).

2. The range is the set of values onto which ƒ sends the x's.

The x's are sent the numbers 1, 2, 3, 1, 7 which ends up looking like {1,2, 3, 7}

So the answer is {1,2,3,7}

You might be interested in

The two triangles are similar. What is the value of x?

denpristay [2]

[8/(2x-2)=(8+6)/(3x)
8/(2x-2)=14/(3x)
14(2x-2)=8(3x)
28x-28=24x
x=7
So by using the similarity of the sides, we can make ratios that can give is the value of x.

3 0

3 years ago

See the attached file.

Masja [62]

f(x) = 4x - x^2

f(4) = 4(4) - 4^2 = 16 - 16 = 0

f(-4) = 4(-4) - (-4)^2 = -16 - 16 = -32

f(4) - f(-4) = 0 - (-32) = 32

f(3/2) = 4(3/2) - (3/2)^2

f(3/2) = 6 - 9/4 = 15/4

√f(3/2) = √(15/4) = √15 / 2

f(x + h) = 4(x + h) - (x + h)^2

= 4x + 4h -(x^2 + 2xh + h^2)

= 4x + 4h -x^2 - 2xh - h^2

f(x - h) = 4(x - h) - (x - h)^2

= 4x - 4h -(x^2 - 2xh + h^2)

= 4x - 4h -x^2 + 2xh - h^2

[f(x + h) -f(x - h) ] / 2h

= [4x + 4h -x^2 - 2xh - h^2 - ( 4x - 4h -x^2 + 2xh - h^2 )] / 2h

=( 4x + 4h -x^2 - 2xh - h^2 - 4x + 4h + x^2 - 2xh + h^2 ) / 2h

= (8h - 4xh) / 2h

= 2h(4 -2x) / 2h

= 4 - 2x

Answer: [f(x + h) -f(x - h) ] / 2h = 4 - 2x

3 0

2 years ago

I don't know how to turn 200 kids into a percent

malfutka [58]

Divide 47 by 100. you will get a decimal. move the decimal two places the right and there is your percent!

7 0

3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of