Multiply polynomials (4x-5)^2

Jessica and her sister have a combined age of 47. Jessica is 5 years older than her sister. How old is Jessica

maw [93]

Let the sister be x years old.
Then Jessica is x + 5 years old.
Their ages add to 47.

x + x + 5 = 47

2x + 5 = 47

2x = 42

x = 21

x + 5 = 26

Jessica is 26, and her sister is 21.

4 0

4 years ago

The data represent the age of world leaders on their day of inauguration. Find the five-number summary, and construct a boxplot

Mumz [18]

Answer:

Minimum value = 43, $Q_1=52$ , $Q_2=Median=61$ , $Q_3=67$ , Maximum value = 69.

Distribution is negative skewed.

Step-by-step explanation:

The given data set is

44, 43, 67, 68, 57, 65, 52, 62, 69, 54, 53, 61, 67, 48, 65

Arrange the data in ascending order

43, 44, 48, 52, 53, 54, 57, 61, 62, 65, 65, 67, 67, 68, 69

Divide the data in 4 equal parts.

(43, 44, 48), 52, (53, 54, 57), 61, (62, 65, 65), 67, (67, 68, 69)

Minimum value = 43

$Q_1=52$

$Q_2=Median=61$

$Q_3=67$

Maximum value = 69

Starting and end point of box plot represent the minimum and maximum value respectively. Box starts from the first quartile to the third quartile. A vertical line goes through the box at the median.

The shape of the distribution.

Normally distributed: If $Q_3-Q_2=Q_2-Q_1$

Positive skew: If $Q_3-Q_2>Q_2-Q_1$

Negative skew: If $Q_3-Q_2$

For the given data set

$Q_3-Q_2=67-61=6$

$Q_2-Q_1=61-52=9$

Since $Q_3-Q_2$ , therefore the given distribution is negative skewed.

5 0

4 years ago

The circle for India has a radius of 3.2 cm and the circle for China has a radius of 3.4 cm. If 17% of the world’s population is

defon

World's population that lives in China in percentage is 20%

8 0

4 years ago

Vita wants to center a towel bar on her door that is 27 1/2 inches wide. She determines that the distance from each end of the t

timurjin [86]

Answer:

The equation is $x+18 =\frac{55}{2}$ ..

The length of the towel bar is $\frac{19}{2}\ in \ \ \ OR \ \ \ 9\frac{1}{2}\ in$ ..

Step-by-step explanation:

Given,

Length of the door = $27\frac{1}{2}\ in$

we can rewrite $27\frac{1}{2}\ in$ as $\frac{55}{2}\ in$ .

We have to find out the length of the towel bar.

Let the length of towel bar be 'x'

`Since Vita wants that the distance from each end of the towel bar to the end of the door to be 9 inches.

So we can say that the length of the towel bar plus from each end of the towel bar to the end of the door plus from each end of the towel bar to the end of the door is equal to length of door.

framing in equation form, we get;

$x+9+9= \frac{55}{2}\\\\x+18 =\frac{55}{2}$

Hence The equation is $x+18 =\frac{55}{2}$ .

Now we solve the equation we get;

Subtracting both side by 18 we get;

$x+18-18=\frac{55}{2}-18\\\\x=\frac{55}{2}-18$

Now we will make the denominator common we get;

$x=\frac{55}{2} -\frac{18\times2}2= \frac{55}{2} -\frac{36}2 = \frac{55-36}{2}=\frac{19}{2}\ in \ \ \ OR \ \ \ 9\frac{1}{2}\ in$

Hence The length of the towel bar is $\frac{19}{2}\ in \ \ \ OR \ \ \ 9\frac{1}{2}\ in$ .

6 0

4 years ago

The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model. The silver model requires 1 minute

tresset_1 [31]

Answer:

Optimal production = 600 gold pens

Revenue = 600*7 = $4200 gold pens

Step-by-step explanation:

The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model.

A. The silver model requires 1 minute in a grinder and 3 minutes in a bonder.

B. The gold model requires 3 minutes in a grinder and 4 minutes in a bonder.

Because of maintenance procedures,

C. the grinder can be operated no more than 30 hours per week and

D. the bonder no more than 50 hours per week.

The company makes

E. $5 on each silver pen and

F. $7 on each gold pen.

How many of each type of pen should be produced and sold each week to maximize profits?

Solution:

We will solve the problem graphically, with number of silver pens, x, on the x axis, and number of gold pens, y, on the y axis, i.e.

1. From A and C, the maximum number of silver pens

x <= 30*60 / 1 = 1800 and

x <= 50*60 /3 = 1000 ....................(1) bonder governs

2. from A & D, the maximum number of gold pens

y <= 30*60 / 3 = 600 .....................(2) grinder governs

y <= 50*60 / 4 = 750

3. From D,

x + 3y <= 30*60 = 1800 ...................(limit of grinder) ..... (3)

3x + 4y <= 50*60 = 3000 .................(limit of bonder) .......(4)

Need to maximize profit,

Z(x,y) = 5x+7y, represented by parallel lines y = -5x/7 + k such that all constraints of (3) and (4) are satisfied.

The maximum is obtained when Z passes through (360,480), i.e. at intersection of constraints (3) and (4). Using slope intercept form,

(y-480) = -(5/7)(x-360)

or y=-(5/7)x + (737+1/7) [the purple line] which violates the red line, so not a solution.

Next try the point (0,600)

(y-600) = -(5/7)(x-0), or

y = 600 - (5/7)x [the black line]

As we can see all point on the black (in the first quadrant) satisfy the constraints, so it is a feasible solution, and is the optimal solution, with a revenue of

Revenue = 600*7 = 4200 gold pens

8 0

4 years ago