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2 years ago

What is the most precise classification of the quadrilateral formed by connecting the midpoints of the sides of a square?

1 answer:

7 0

A square again
Lets take a is side of a square.
if u take parts of square from mid point and measure distance or length of any sides that would be a/2

then use Pythagorean theorem to solve hypotenuse u will get all length of small quadrilateral a/√2.same for all sides.
I see that will be square again

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Connie has saved up $15 to purchase a new CD from the local store. The sales tax in her county is 5% of the sticker price. Write

Sholpan [36]

Answer:

$x + 0.05x = 15$ , solution is $14.29

Step-by-step explanation:

We can create an equation for this scenario to try and solve it.

Assuming the cost of the CD is x, it’s sale tax will be 0.05x (as that is 5% of x, 0.05.)

We can write the equation in two ways:

$x + 0.05x = 15$ or $1.05x = 15$ .

Assuming we take the second one (easier to work with), we can divide both sides by 1.05. The equation simplifies to x = 14.289... which rounds to x = 14.29.

Therefore, the equation is $x + 0.05x = 15$ and the solution is $14.29.

Hope this helped!

3 0

3 years ago

Read 2 more answers

Plot the following ordered pairs on a graph and then determine whether or not the points represent a function.

Alex777 [14]

The answer is :

YES: These points represent a function because none of the vertical lines I drew passed through more than one point.

5 0

3 years ago

Read 2 more answers

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$