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3 years ago

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sally needs $ 20,000 in 4 years, she has $ 10,000 to invest. if interest is compounded continuously, what rate is required for s

ally to meet her goal?

1 answer:

Amiraneli [1.4K]3 years ago

3 0

She needs to save 5,000 per year, that way she can reach 20,000 in 4 years

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Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can

dimaraw [331]

Answer:

(a) 5 senior, 4 apprentice

(b) 368 per shift

(c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop

5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves

These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

(x, y) = (0, 7.5), (4, 5), (9, 0)

__

(a) Santa wants to maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

t(0, 7.5) = 45

t(4, 5) = 46

t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

__

(b) The above calculations show 46 trucks per hour can be made, so ...

46×8 = 368 . . . trucks in an 8-hour shift

__

(c) The new demands change the inequalities to ...

x + y ≤ 8 . . . . . . number of workers

7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

(x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

_____

<em>Comment on apprentice elf wages</em>

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.

5 0

3 years ago

If: \qquad AC = 64 AB = 7x + 8, and BC= 3x+6, Find BC

Nady [450]

The answer is 21.

First you Add both AB and BC which is 10x +14. Then you put 64 into the equation. 10x+14=64. Subtract 14 from 64 which is 50. So 10x=50. Then divide 10 from 50. which would be x=5. So if you plug it in. It would be 21.

4 0

3 years ago

A study was conducted to test whether a manufacturing process performed at a remote location can be established locally. Test de

zhenek [66]

Answer:

Step-by-step explanation:

Hello!

You have the information about voltage readings at an old and a new manufacturing location obtained remotely.

a, b and c in the attachment.

Histograms for a and c:

To construct a frequency histogram you have to first arrange the data for both locations in a frequency table. For this, I'm going to determine 5 class interval for each location. To do so you need to calculate the width of the intervals. First, you calculate the range of the variable and then you have to divide it by the number of intervals you want to do.

Old location: Range= 10,55-8,05= 2,5 → Class width: 2,5/5= 0,5

New Location: Range= 10,12-8,51= 1,61 → Class width: 1,61/5= 0,322

Starting from the minimum value you add the calculated width and create the intervals:

Old Location:

8,05-8,55

8,55-9,05

9,05-9,55

9,55-10,05

10,05-10,55

New Location

8,51-8,83

8,83-9,15

9,15-9,48

9,48-9,80

9,80-10,12

Stem and Leaf diagram for b:

To construct this diagram first I've ordered the data from leat to greatests. Then I've used the integer to form the stem 8,- 9.- and 10.- and the decimals are placed in the leafs of the diagram.

Comparing it to the histogram and stem and leaf diagram for the readings of the Old Location, the histogram stem, and leaf diagram show better where most of the readings lie.

d.

Comparing both histograms, it looks like the readings in the new location are more variable than the readings in the old location but more uniformly distributed. I would say that the readings in the new location are better than the readings in the old location.

e.

To calculate the mean you have to apply the following formula:

X[bar]= (∑xi'fi)/n

X[bar]OLD=(∑xi'fi)/n= (8.3*1+8.8*3+9.3*0+9.8*17+10.3*9)/30= 294/30= 9.8

X[bar]NEW=(∑xi'fi)/n= (8.67*6+8.99*2+9.315*7+9.64*8+9.96*7)/30= 282.045/30= 9.4015≅9.40

First you have to calculate the position of the median:

For both data sets the PosMe= 30/2=15

Now you arrange the data from least to highest and determine wich observation is in the 15th position:

Old Location

8,05 , 8,72 , 8,72 , 8,8 , 9,55 , 9,7 , 9,73 , 9,8 , 9,8 , 9,84 , 9,84 , 9,87 , 9,87 , 9,95 , 9,97 , 9,98 , 9,98 , 10 , 10,01 , 10,02 , 10,03 , 10,05 , 10,05 , 10,12 , 10,15 , 10,15 , 10,26 , 10,26 , 10,29 , 10,55

MeOLD= 9.97

New Location

8,51 , 8,65 , 8,68 , 8,78 , 8,82 , 8,82 , 8,83 , 9,14 , 9,19 , 9,27 , 9,35 , 9,36 , 9,37 , 9,39 , 9,43 , 9,48 , 9,49 , 9,54 , 9,6 , 9,63 , 9,64 , 9,7 , 9,75 , 9,85 , 10,01 , 10,03 , 10,05 , 10,09 , 10,1 , 10,12

MeNEW= 9.43

The mode is the observation with more absolute frequency.

To determine the mode on both data sets I'll use the followinf formula:

Md= Li + c [Δ₁/(Δ₁+Δ₂)]

Li= Lower bond of the interval with most absolute frequency (modal interval)

c= amplitude of the modal interval

Δ₁= absolute frequency of the modal interval minus abolute frequency of the previous interval

Δ₂= absolute frequency of the modal interval minus the absolute frequency of the next interval

Modal interval OLD

9,55-10,05

Δ₁= 17-0= 17

Δ₂= 17-9= 8

c= 0.5

Li= 9.55

MdOLD= 9.55 + 0.5*[17/(17+8)]= 9.89

Modal interval NEW

9,48-9,80

Δ₁= 8-7= 1

Δ₂= 8-7= 1

c= 0.32

Li= 9.48

MdNEW= 9.48+0.32*[1/(1+1)]= 9.64

f.

OLD

Mean 9.8

SE 0.45

X= 10.50

Z= (10.50-9.8)/0.45= 1.56

g.

NEW

Mean 9.4

SE 0.48

X=10.50

Z= (10.50-9.4)/0.48=2.29

h. The Z score for the reading 10.50 for the old location is less than the Z score for the reading 10.50 for the new location, this means that the reading is closer to the mean in the old location than in the new location.

The reading 10.50 is more unusual for the new location.

i. and k. Boxplots attached.

There are outliers for the readings in the old location, none in the readings for the new location.

j. To detect outliers using the Z- score you have to "standardize every value of the data set using the corresponding values of the mean and standard deviation. Observations that obtained a Z-score greater than 3 or less than -3 are outliers.

The data set for the new location has no outliers, To prove it I've calculated the Z-scores for the max and min values:

Min: Z=(8.51-9.4)/0.48= -1.85

Max: Z=(10.12-9.4)/0.48= 1.5

The records for the old location show, as seen in the boxplot, outliers:

To find them I'll start calculating values of Z from the bottom and the top of the list until getting a value Z≥-3 and Z≤3

Bottom:

1) 8,05 ⇒ Z=(8.05-9.8)/0.45= -3.89

2) 8,72 ⇒ Z= (8.72-9.8)/0.45= -2.4

Top

1) 10,55⇒ Z= (10.55-9.8)/0.45= 1.67

m. As mentioned before, the distribution for the new location seems to be more uniform and better distributed than the distribution for the old location. Both distributions are left-skewed, the distribution for the data of the old location is severely affected by the presence of outliers.

I hope this helps!

5 0

3 years ago

The linear equation y = 30x describes how far from home Traci is as she drives from Dallas to Anchorage. Let x represent the num

Alex777 [14]

Traci will be 360 miles in 12 hours if the linear equation y = 30x describes how far from home Traci is as she drives from Dallas to Anchorage and the equation of the graph is linear.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have a linear equation:

y = 30x

Here, x represent the number of hours and y represent the number of miles.

When x = 12 hours then

y = 30(12) = 360 miles

If draw the graph the equation y = 30x, it will be a linear equation.

(refer attached picture)

Thus, Traci will be 360 miles in 12 hours if the linear equation y = 30x describes how far from home Traci is as she drives from Dallas to Anchorage and the equation of the graph is linear.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

3 0

2 years ago

Fill in the missing term in the equation. x/x^2-4x+4-x/x^2-3x+2=?/(x-2)^2(x-1)

emmainna [20.7K]

<span> x x
----------------- - --------------------
x^2 -4x + 4 </span><span>x^2-3x+2

</span> x x
= ----------------  - --------------------
(x -2)^2 (x - 2)(x - 1)

x (x-1) - x(x +2)
= ----------------------------
(x -2)^2 (x - 1)

x^2 -x -x^2 - 2x
= --------------------------
    (x -2)^2 (x - 1)

-3x
= -----------------------
    (x -2)^2 (x - 1)

answer
<span>missing term in the equation: -3x</span>

5 0

3 years ago

Read 2 more answers