All categories

3 years ago

The selling price of a television in a retail store is $66 less than 3 times the wholesale price. If the selling price

Mathematics

1 answer:

Anna71 [15]3 years ago

5 0

Answer:

899 + 66 = 3*X, X being the wholesale price.

965 = 3*x -> x = 965/3 -> x = 321.66 (and 2/3 of a cent)

Step-by-step explanation:

You might be interested in

Consider the following information related to a set of quantitative sample data:

svlad2 [7]

Answer:

(a) There are outliers

(b) $x$ and $x >62$

Step-by-step explanation:

Given

$\sigma = 14.92$

$\bar x = 22.0$

$q_0 = -24$

$q_1 = 14.5$

$q_2 = 24.5$

$q_3 = 33.5$

$q_4 = 64$

Solving (a): Check for outliers

This is calculated using:

$Lower = Q_1 - (1.5 * IQR)$ --- lower bound of outlier

$Upper = Q_3 +(1.5 * IQR)$ --- upper bound of outlier

Where

$IQR = Q_3 - Q_1$

So, we have:

$IQR = 33.5 - 14.5$

$IQR = 19$

The lower bound of outlier becomes

$Lower = Q_1 - (1.5 * IQR)$

$Lower = 14.5 - (1.5 * 19)$

$Lower = 14.5 - 28.5$

$Lower = -14$

The upper bound of outlier becomes

$Upper = Q_3 +(1.5 * IQR)$

$Upper = 33.5 + 1.5 * 19$

$Upper = 33.5 + 28.5$

$Upper = 62$

So, we have:

$-14 \le x \le 62$ --- the range without outlier

Given that:

$q_0 = -24$ --- This represents the lowest data

$q_4 = 64$ --- This represents the highest data

-24 and 64 are out of range of $-14 \le x \le 62$ .

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

$x$ and $x >62$

4 0

3 years ago

Which term is a perfect square of the root 3x4?

hodyreva [135]

Answer:

Step-by-step explanation:

because eight plus five is 85

3 0

3 years ago

Read 2 more answers

How do I change my profile picture, plz answer.

zaharov [31]

Answer:

got to settings

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago

Find the perimeter. Explain what you added together to get your answer. You can get partial credit for explaining steps, even if

sertanlavr [38]

Answer:

the area of the shape is 7.14

Step-by-step explanation:

First, find the area of the square:

2 in x 2 in = 4 in

then to find the area of the semi- circles add them together to make a circle

To find the area of the circle, use:

pi x radius^2

Since the diameter is 2 divide the diameter by 2 to get radius

3.14 x 1^2 = 3.14

The circle is 3.14 plus 4 ( the square ) = 7.14

5 0

3 years ago