LITERAL EQUATIONS K= AP/2 SOLVE FOR A

1 answer:

4 0

$K=\frac{AP}{2}\ \ \ \ |Multiply\ by\ 2\\\\
2K=AP \ \ \ \ \ |Divide\ by\ P\\\\
\frac{2K}{P}=A\\\\Solution\ is\ A=\frac{2K}{P}.$

You might be interested in

3) The foot of a ladder is 1.2 m from a fence

VashaNatasha [74]

Answer: $4.5\ meters$

Step-by-step explanation:

Using the data given in the exercise, we can draw the diagram attached, where "h" is the height of the building reached by the top of the ladder.

Notice that there are two similar triangles.

So, you can set up the following proportion:

$\frac{1.8}{1.2}=\frac{h}{(1.2+1.8)}$

Finally, in order to calculate the height on the building reached by the top of the ladder, you must solve for "h".

Therefore, the value of "h" is :

$\frac{1.8}{1.2}=\frac{h}{3}\\\\(3)(\frac{1.8}{1.2})=h\\\\\frac{5.4}{1.2}=h\\\\h=4.5$

5 0

3 years ago

The solutions to the inequality y<-x+1 are shaded on the graph. Which point is a solution?

suter [353]

Answer:
B) (3, –2)

Explanation:
The inequality is y ≤ –x + 1
There are two ways to do this. You can try the four options by seeing where they lie on the graph, or by inputting them into the inequality and seeing if they check out. I am going to do a bit of both.

I know that the solution cannot have two positive coordinates because the first quadrant is not part of the solution, so I won't guess A or C.
I'll try (3, –2) (which is option B).
On the graph, (3, –2) is on the line, which means it is part of the solution because the line is solid and the inequality is a greater than or equal to sign.
Try it in the inequality:
y ≤ –x + 1
–2 ≤ –3 + 1
–2 ≤ –2 yes this checks out.

9 0

3 years ago

I need help with 0.97 of 1/10 of

Svetach [21]

Its .097. Just move the decimal

8 0

3 years ago

A simple random sample of 100 observations was taken from a large population. The sample mean and the standard deviation were de

olga2289 [7]

Answer:

Se=1.2

Step-by-step explanation:

The standard error is the standard deviation of a sample population. "It measures the accuracy with which a sample represents a population".

The central limit theorem (CLT) states "that the distribution of sample means approximates a normal distribution, as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population distribution shape"

The sample mean is defined as:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

And the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

Let X denotes the random variable that measures the particular characteristic of interest. Let, X1, X2, …, Xn be the values of the random variable for the n units of the sample.

As the sample size is large,(>30) it can be assumed that the distribution is normal. The standard error of the sample mean X bar is given by:

$Se=\frac{\sigma}{\sqrt{n}}$