Graphically, a point is a solution to a system of two inequalities if and only if the point

Ivanshal [37]

Answer:

the answer is median (b)

8 0

3 years ago

A rectangular prism has a base that is 5 cm by 7 cm and a height of 12 cm. if all dimensions are doubled, what happens to the vo

Maurinko [17]

We know, Volume of rectangular prism = l * b * h
If all dimensions are doubled, then it would be: 2 * 2 * 2 * v
V' = 8V

In Short, Volume will increase to it's 8 times

Hope this helps!

7 0

3 years ago

Read 2 more answers

Find the slope and y-intercept in the equation.<br> 2x+4y=9

const2013 [10]

The first thing to do with these problems is to convert them into y=mx+b form; this will literally show you the answer to slope and y intercept since m=slope and b=y intercept.

To convert, isolate the y and the rest should fall into place. Subtract 2x from 4y and the other side of the equation to get the equation 4y= -2x+9 (put the x first because of mx+b). Then Divide the 4 from y and do the same to the other side to get y= -1/2x+9/4. M= -1/2 so that is the slope. B=9/4 so that is the y intercept

4 0

4 years ago

Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,

Bas_tet [7]

Answer:

a) $\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

$Median = 25$

b) $Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) $Midrange = \frac{70+13}{3}=41.5$

d) $Q_1 = \frac{20+21}{2} =20.5$

$Q_3 =\frac{35+35}{2}=35$

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

$\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96$

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

$Median = 25$

Part b

The mode is the most repeated value on the dataset on this case would be:

$Mode = 25, 35$

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

$Midrange = \frac{Max+Min}{2}$

And if we replace we got:

$Midrange = \frac{70+13}{3}=41.5$

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

$Q_1 = \frac{20+21}{2} =20.5$

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

$Q_3 =\frac{35+35}{2}=35$

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0

4 years ago

A business has two loans totaling $50,000. One loan has a rate of 8% and the other has a rate of 12%. This year, the business ex

Debora [2.8K]

Answer:

the 8% loan has a principal of $37500

the 12% loan has a principal of $12500

Step-by-step explanation:

Let's start by writing the general equation for the interest hwre I is the interest, P is the principal (in our case would be loan amounts), "r" is the interest rate in decimal form (in our case one would be 0.12, and the other one 0.08), and t is the time in years (in our case 1 year).

$I=P*r*t$

Then we write the interest equation coming from each loan at the end of this year (we call I1 the interest coming from the 12% loan and I2 the interest coming from the 8% one). Since we don't know the loan amounts (in fact those are what we need to find) we will name one "x" and the other "y":

$I=P*r*t\\I1=x * 0.12*1\\I2=y*0.08*1$

Next, we add these last two equations term by term, and replace the addition of both interests by $4500 as given in the information:

$I1=x * 0.12*1\\I2=y*0.08*1\\I1+I2 = 0.12x+0.08y\\4500=0.12x+0.08y$

This is our first equation in the variables x and y which are our unknowns.

Now we generate the second equation on x and y by writing in agebraic terms the other piece of information we have: "the total of the two loans is $50000. That is the addition of the principals x and y should equal $50000:

$x+y=50000$

We solve for y in this last equation and replace its form in terms of x in the equation of the interest, and solve for the unknown x:

$y=50000-x\\4500 = 0.12x +0.08 y\\4500=0.12x+0.08(50000-x)\\4500=0.12x+4000-0.08x\\4500=0.12x-0.08x+4000\\4500=0.04x+4000\\4500-4000=0.04x\\500=0.04x\\x=\frac{500}{0.04} =12500$

Therefore the amount of the loan at 12% is $12500

Now to find the amount of the second loan "y" we use the equation for the totals of the loans:

$x+y=50000\\12500+y=50000\\y=50000-12500=37500$

Therefore, the loan at 8% is $37500

5 0

3 years ago