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frozen [14]
4 years ago
11

A+12/3=8 what is the answer to this math problem

Mathematics
2 answers:
melomori [17]4 years ago
8 0

Answer:

a=4

Step-by-step explanation:

a+12/3=8

a+4=8

a=8-4

a=4

Nadusha1986 [10]4 years ago
7 0

Answer:

4

Step-by-step explanation:

a + 12/3 = 8

a + 4 = 8

a = 8 - 4

a = 4

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What is the slope-intercept form of linear equations?
lara31 [8.8K]

to get the equation of any straight line, we simply need two points off of it, let's use the points from the picture below.

(\stackrel{x_1}{8}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{12}-\underset{x_1}{8}}}\implies \cfrac{2}{4}\implies \cfrac{1}{2}

\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-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{8}) \\\\\\ y-3=\cfrac{1}{2}x-4\implies y=\cfrac{1}{2}x-1

if we already have the slope, and we can see the y-intercept on the table, then we can simply use the slopel-intercept form and plug both of them in.

\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \begin{array}{|c|ll} \cline{1-1} slope\\ \cline{1-1} \cfrac{1}{2}\\ \cline{1-1} y-intercept&\\\cline{1-1} (0~~,~~-1)\\ \cline{1-1} \end{array}~\hfill y=\cfrac{1}{2}x-1

6 0
2 years ago
Find the critical value needed to construct a 95% CI of a population mean with an unknown standard deviation and a sample size o
kaheart [24]

Answer:

t_{\alpha/2}=-2.02, t_{1-\alpha/2}=2.02

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

In order to find the critical value we need to take in count that we are finding the critical value for the population mean and without the standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. We need to calculate the degrees of freedom are given by:

df=n-1=39-1=38

We can find the critical value using the following excel codes:

"=T.INV(0.025;38)" or "T.INV(1-0.025;38)"

And the critical values would be given by:

t_{\alpha/2}=-2.02, t_{1-\alpha/2}=2.02

7 0
3 years ago
Determine the y-intercept and the equation for the horizontal asymptote of the function.
makkiz [27]
K(x) = (1/2)ˣ
Since a (=1/2) is < 1, this exponential function is decreasing .
 y-intercept for x = 0
 k(x) =(1/2)⁰  → k(x) =1, then y-intercept = 1

Horizontal Asymptote:

lim k(x) = (1/2)ˣ = 0
x→∞

Horizontal asymptote x = 0

8 0
3 years ago
-16.5v − 19.94 = 19.39 − 9.6v
kramer

Answer:


Step-by-step explanation:

Hi there,

To solve this problem, we will need to isolate the variable by moving the variable to one side and the number to the other side.

1. - 16.5v - 19.94 = 19.39 - 9.6v <em>Rewrite the problem</em>

2. - 16.5v = 39.33 - 9.6v <em>Move the non-variable number to one side</em>

3. -6.9v = 39.33 <em>Isolate the variable</em>

4. v = - 5.7 <em>Divide by -6.9</em>

Thus, we can see the answer to this problem is v = - 5.7.

Have a good day,

Jamie

3 0
3 years ago
PLEASE HELP!!
saw5 [17]
(0.4×30,000)−5,000
=7,000
4 0
3 years ago
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