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3 years ago

Can you write this equation in point-slope form? y = ½x + 4?

1 answer:

4 0

Yes. This equation given:
______________________________
   " y = (½)x + 4 " ; in point-slope form; also known as: "slope-intercept form" ; is:
______________________________________
    " y = (½)x + 4 " .
______________________________________
In other words, the equation given is ALREADY written in "point-slope form" ; or, "slope-intercept form".
______________________________________
Note:   An equation that is written in "point-slope form"
             (or, "slope-intercept form"), is written in the format of:
______________________________________
" y = mx + b " ;_________________
in which:_________________
"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:
"m" is the slope of the line; which is what we want to solve for;
"b" is the "y-intercept"; or more precisely, the value of "x"
(that is; the "x-coordinate") of the point at which "y = 0";
that is, the value of "x" ; or the "x-coordinate" of the point at which
the graph of the equation crosses the "x-axis".
______________________________________
Note that in our given equation, which is written in "point-slope form" (or, "slope-intercept form" — that is: " y = mx + b " ;
_______________________________________
which is: " y = (½)x + 4 " ;
_______________________________________
we have:
_______________________________________
"y" isolated as "stand-alone" variable on the "left-hand side" of the equation;

m = ½ ;
b = 4 .
_______________________________________

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Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% 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 age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

Solve y + 4 > - 2.2. Check your solution.

maks197457 [2]

Pretty sure its y > -6.2
This is because you subtract 4 from both sides and get y > -6.2
Please correct me if im wrong thx!

6 0

3 years ago

Read 2 more answers

K(x)=-3x^2-3;find k(-2)

Agata [3.3K]

Answer:

Step-by-step explanation:

because I leaned it at school

6 0

2 years ago

The area of a 12-cm-wide rectangle is 288 cm^2 what is its length?

densk [106]

Answer:

24cm

Step-by-step explanation:

The area of the rectangle is length × breadth. You already know the area and the breadth, so you can find the length.

288cm² = length × 12cm

length = 288cm² ÷ 12cm

= 24cm

3 0

3 years ago

If f(x) = x-1/3 and g(x)= 3x+1, what is (f o g)(x)?

N76 [4]

Answer:

$(f o g)(x) = 3x + \frac{2}{3}$

Step-by-step explanation:

We have the function $f(x) = x-\frac{1}{3}$ and we have the function $g(x) = 3x + 1$ . We want to find g(x) composed with f(x)

Then, the function (f o g)(x) is the same since f(g(x))

That is, you must do x = g(x) and then enter g(x) into the function f(x).

$f(g(x)) = (g(x)) -\frac{1}{3}$

$f(g(x)) = (3x + 1) -\frac{1}{3}$

Simplifying, we obtain:

$(f o g)(x) = 3x + 1 -\frac{1}{3}\\\\(f o g)(x) = 3x + \frac{2}{3}$

Finally. The composite function is:

$(f o g)(x) = 3x + \frac{2}{3}$

6 0

3 years ago