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3 years ago

Find the slope and y-intercept of the line. 18x + 4y = 112

2 answers:

4 0

To find the slope and y-intercept, you can convert the linear equation into the form y = mx + b, where m is the slope and b is the intercept.
To do so, just simplify for y.
$18x+4y=112$
$4y=-18x+112$
$y = - \frac{18}{4}x+28$
So, the y-intercept is (0, 28) and the slope is 4.5.
Hope that helps!

nadezda [96]3 years ago

3 0

Answer:

slope: -9/2

y intercept: 28

Step-by-step explanation:

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Slope of line y = -2x+ 4

Fofino [41]

Answer:

-2

Step-by-step explanation:

The equation of a line in slope intercept form is

y = mx +b where m is the slope and b is the y intercept

y = -2x+4

The slope is -2 and the y intercept is 4

3 0

3 years ago

Calculate the total amount and compound interest on RS 8,000 for a year at the rate of 9% per annum compounded half yearly.

Gennadij [26K]

Answer:

See below in bold.

Step-by-step explanation:

Total = A(1 + r/n)^nt where A = initial amount , t = number of years, r = the rate (as a decimal fraction) and n = number of payments a year. So:

Total after 1 year = 8000(1 + 0.09/2) ^ 1*2

= RS 8736.20.

Compound interest = RS 736.20

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3 years ago

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n

Svetradugi [14.3K]

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

The null hypothesis is $H_o : p = 0.015$

The alternative hypothesis is $H_a : p < 0.015$

The sample size is n= 150

Generally in order to use normal sampling distribution

The value $np \ge 5$

$np = 0.015 * 150$

$np = 2.25$

Given that $np < 5$ normal sampling distribution can not be used

8 0

3 years ago

What is the general form of the equation of the line shown?

ZanzabumX [31]

Check the picture below. So let's use those two points on the line.

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-0}{3-0}\implies \cfrac{3}{3}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=1(x-0)\implies y=x \\\\\\ -x+y=0\implies \stackrel{\textit{standard form}}{x-y=0}$