Suppose Diana Places $4500 In An Account That Pays 8% Interest Compounded Each Year. Assume That No Withdrawals (2024)

The percent error is given by:

[tex]E=\frac{|measured-real|}{real}\times100[/tex]

For trial A the percent of error is:

[tex]E=\frac{|240-200|}{200}\times100=20[/tex]

20%

For trial B, the percent of error is:

[tex]E=\frac{|195-240|}{240}\times100=18.75[/tex]

18.75%

Given:

Here we have z=1.3 and z=3.

Required:

We need to find percentage of data items in a normal distribution that lie between z=1.3 and z=3.

Explanation:

We have z=1.3 and the percentage of z=1.3 is 90.32 % and percentage of z=3 is 99.87 %

now to find between these two values we just need subtract both values

[tex]99.87-90.32=9.55\text{ \%}[/tex]

Final Answer:

9.55 %

The company's annual profit (in millions of dollars) as a function of the app's price in dollars) is modeled by :

P(x) = -2(x - 3)(x - 11)

If the app price is zero : i.e. x = 0

Susbtitute x = 0 in the function of P(x)

P( x ) = -2( x - 3 )( x - 11 )

P( 0 ) = -2( 0 - 3 )( 0 - 11 )

P ( 0 ) = -2(-3)( -11)

P( 0 ) = 6( - 11 )

P( 0 ) = -66

There will be the loss of 66 dollars

Answer : -66 million dollars

f(x) = -2x - 3

Substituting with x = 3, we get:

f(3) = -2(3) - 3

f(3) = -6 - 3

f(3) = -9

hello

to find the constant of proportionality, we should use the formula given

[tex]\begin{gathered} y=rx \\ \end{gathered}[/tex][tex]\begin{gathered} y_1=36 \\ x_1=12 \\ y=rx \\ r=\frac{y}{x} \\ r=\frac{36}{12} \\ r=3 \end{gathered}[/tex]

let's check for another point

[tex]\begin{gathered} y_2=54 \\ x_2=18 \\ y=rx \\ r=\frac{y}{x} \\ r=\frac{54}{18} \\ r=3 \end{gathered}[/tex]

from the calculations above, the value of r is equal to 3

Given:

The maximum speed (v) that an automobile can travel around a curve of radius r without skidding is given by the equation:

[tex]v=(\frac{5r}{2})^{\frac{1}{2}}[/tex]

We will find the maximum speed v when r = 40 feet

so, substitute r = 40

[tex]v=(\frac{5*40}{2})^{\frac{1}{2}}=(100)^{\frac{1}{2}}=10[/tex]

So, the answer will be:

The car can travel up to 10 miles per hour.

In the given problem the line DE passes through the center of the circle hence the line DE is the diameter of the circle.

From the property of the circle the angle subtended by the diameter on the circle is a right angle.

Hence angle DFE is 90 degrees.

In triangle DFE,

[tex]\begin{gathered} \angle DFE+\angle EDF+\angle DEF=180^{\circ} \\ 90^{\circ}+\angle EDF+63^{\circ}=180^{\circ} \\ \angle EDF=180^{\circ}-153^{\circ} \\ \angle EDF=27^{\circ} \end{gathered}[/tex]

Thus, the arc FE subtends 27 degrees at the circle.

For the Step 1: Find the straight time pay.

The statement tells you that Alonso is paid $ 18.2 an hour for a regular workweek. Also, they tell you that Alonso worked his regular 40 hours this week. So,

[tex]\begin{gathered} \text{Hourly Rate X Regular Hours Worked = Total Straight Time Pay} \\ \text{ \$}18.2\cdot40=\text{ \$}728 \end{gathered}[/tex]

Now for the step 2: Find the overtime pay.

The statement tells you that Alonso's overtime rate is 1.5 times his hourly rate. Then you multiply

[tex]\text{ \$}18.2\cdot1.5=\text{ \$}27.3[/tex]

This means that Alonso is paid $ 27.3 per overtime hour worked. So,

[tex]\begin{gathered} \text{ Overtime Rate X Regular Hours Worked = Total Overtime Pay} \\ \text{ \$}27.3\cdot10=\text{\$}273 \\ \text{ because he worked 10 overtime hours} \end{gathered}[/tex]

Finally, for the step 3: Find the straight time pay.

Now, you just have to add what Alonso earned for his 40 regular hours plus what he earned for his 10 overtime hours.

[tex]\begin{gathered} \text{ Striaght Time Pay + Overtime Pay = Total Amount Earned} \\ \text{ \$}728+\text{\$}273=\text{\$}1001 \end{gathered}[/tex]

That means that this week Alonso made a total of $ 1001.

Answer:

A.) The new mean will be 61.75

Explanation

First we need to find the mean and median of the original data

Given the datas

64, 58, 55, 25, 70

Mean = Sum of data/sample size

Mean = 64+58+55+25+70/5

Mean = 272/5

Mean = 54.4

For the median

Rearrange in ascending order first

25, 55), 58, (64, 70

The number at the middle will be the median

Median = 58

If Ken's value is excluded from the data sample, the remaining data will be;

64, 58, 55, 70

Calculate the new mean

New Mean = 64+58+55+70/4

New Mean = 61.75

Hence if Ken's value is excluded from the data sample, the new mean will be 61.75

Mean Difference = 61.75 - 54.4 = 7.35

This shows that the mean increases if Kens value is excluded

For the new median;

64, 58, 55, 70

New median = 58+55/2 = 56.5

This shows that New median decreases.

Given the linear equation

[tex]2y+3=0[/tex]

To begin with, let us make y the subject of the question

[tex]\begin{gathered} 2y=0-3 \\ 2y=-3 \\ \\ y=-\frac{3}{2} \end{gathered}[/tex]

The graph of the function is given below

The points on the graph include

[tex](0,-1.5)\text{ and (3,-1.5)}[/tex]

For fraction we will have

[tex](0,-\frac{3}{2})\text{ and (}3,-\frac{3}{2}\text{)}[/tex]

The volume ot the given triangular prims is equal to the product of the triangular base and the lenght of 8 meters. The area of the triangular base (A) is given by

[tex]\begin{gathered} A=\frac{base\times height}{2} \\ A=\frac{3\times2}{2} \\ A=3m^2 \end{gathered}[/tex]

Then, the volume of our triangular prisms is given by

[tex]\begin{gathered} V=A\times length \\ V=3\times8 \\ V=24m^3 \end{gathered}[/tex]

Therefore, the volume is equal to 24 cubic meters.

Solution

Dependent events in probability means events whose occurrence of one affect the probability of occurrence of the other. For example suppose a bag has 3 red and 6 green balls. Two balls are drawn from the bag one after the other.

The probability of picking the first card cant affect the probability of picking the second card, hence it is dependent

So

To get the probability of Point P be placed within the circle L, we need to get the ratio between the area covered by the circle and the area covered by the rectangle.

The area of the circle is:

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The temperature of a city, in degrees Fahrenheit, was recorded over a 10-hour period. Consider the function below that represents the relationship between the temperature and the number of hours, x.



Find f(5) and determine what it represents​

__________________________________

f(5)

First, identify the correct function part

As x= 5

4 < x ≤ 6

So we are going to replace the second function part

f(x) =3x + 48

f(5)= 3*5 +48

f(5) = 15 + 48

f(x) = 63

_________________

Answer

f(x) = 63 ; The temperateru

We have, first we need to have all in the same units

1 yard = 36 in

5 yards =180in

1 min =60 sec

12 min=720 sec

With this information we have

180 in -- 720 seconds

? 30 seconds

In order to know the distance in inches in 30 sec

[tex]\frac{30\cdot180}{720}=7.5\text{ in}[/tex]

The distance in 30 seconds is 7.5 inches

Given the following points that pass through a graph.

Point 1: (6, 2)

Point 2: (8, 8)

Let's determine the equation of the line.

Step 1: Determine the slope.

[tex]\text{ Slope = m = }\frac{y_2\text{ - y}_1}{x_2\text{ - x}_1}[/tex][tex]\text{ = }\frac{8\text{ - 2}}{8\text{ - 6}}\text{ = }\frac{6}{2}[/tex][tex]\text{ Slope = m = 3}[/tex]

Step 2: Let's determine the y-intercept (b). Plugin m = 3 and x,y = 6, 2 in y = mx + b.

[tex]\text{ y = mx + b}[/tex][tex]\text{ 2 = \lparen3\rparen\lparen6\rparen + b}[/tex][tex]\text{ 2 = 18 + b}[/tex][tex]\text{ b = 2 - 18 = -16}[/tex]

Step 3: Let's complete the equation. Plugin m = 3 and b = -16 in y = mx + b.

[tex]\text{ y = mx + b}[/tex][tex]\text{ y = \lparen3\rparen x + \lparen-16\rparen}[/tex][tex]\text{ y = 3x - 16}[/tex]

Therefore, the equation of the line is y = 3x - 16

4(k-2) = 2(k-9)



open the parentheses

4k - 8 = 2k - 18

subtract 2k from both-side of the equation

4k - 2k - 8 = 2k - 2k - 18

2k - 8 = -18

Add 8 to both-side of the equation

2k - 8 + 8 = -18 + 8

2k = -10

Divide both-side of the equation by 2

k = -10/2

k= -5

Solve for all values of x in the simplest form.

-|4x + 3] + 3 = -6​

__________________

Can you see the updates?

_________________________

ok

_______________

-|4x + 3] + 3 = -6​

-4x (-3 + 3) = -6​

-4x = -6

x= 6/4

x= 3/2

______________

Answer

x= 3/2

____________

Do you have any questions regarding the solution?

we have that

The total Nolan's spin is equal to

(13+16+9)=38 spins

probability of landing on red

P=13/38

probability of landing on yellow

P=16/38

simplify

P=8/19

probability of landing on blue

P=9/38

[tex]-\frac{2}{3}x+7<15[/tex]

multiply both sides of the inequality by 3

[tex]\begin{gathered} 3\cdot(-\frac{2}{3}x+7)<15\cdot3 \\ -2x+21<45 \end{gathered}[/tex]

bring all constants to the right of the inequality

[tex]\begin{gathered} -2x<45-21 \\ -2x<24 \end{gathered}[/tex]

divide by -2 both sides of the inequality and invert the sign

[tex]\begin{gathered} x>\frac{24}{-2} \\ x>-12 \end{gathered}[/tex]

ANSWER:

2001-2002

Absolute increase: 0.43 trilion

Relative increase: 7.45%

2005-2006

Absolute increase: 0.54 trilion

Relative increase: 6.83%

In the period 2001-2002 it is the larger increase in federal debt as a percentage and in terms of money, the larger increase was the period 2005-2006.

STEP-BY-STEP EXPLANATION:

The absolute increase is the difference between both values and the relative increase would be the percentage of increase between the values.

Therefore, we calculate both increments in both cases:

Period 2001-2002 (in trillions):

[tex]\begin{gathered} I_A=6.20-5.77=0.43 \\ I_R=\frac{6.20-5.77}{5.77}\cdot100=7.45\text{\%} \end{gathered}[/tex]

Period 2005-2006 (in trillions):

[tex]\begin{gathered} I_A=8.45-7.91=0.54 \\ I_R=\frac{8.45-7.91}{7.91}\cdot100=6.83\text{\%} \end{gathered}[/tex]

Therefore in the period 2001-2002 the absolute increase is 0.43 trillion and in the period 2005-2006 it is 0.54 trillion.

But the relative increase is 7.45% and 6.83%, respectively, therefore we can say that the larger increase occurred in the period 2001-2002 percentage and if we take into account the numerical value, the value is greater in the period 2005-2006, therefore in terms of money the increase is larger in this period

1 quart = 0.25 gallon

6 quarts = x

x = 6 (0.25) = 1.5

Therefore,

[tex]6\text{ quarts = 1}\frac{1}{2}\text{ gallons}[/tex]

An equation without solutions is an equality which gives a false result like 1=0 or 3=4.

In this case, the first equation has no solution, let's prove it

[tex]\begin{gathered} 4(x+2)=4(x+3) \\ 4x+8=4x+12 \\ 8=12 \end{gathered}[/tex]

As you can observe, the solution is not true which means the equation has no solutions at all.

Therefore, the answer is the first equation.

Solution

we have the following data:

Year Users

2001 155

2004 190

2007 220

2009 230

And we want to find the slope between 2007 and 2009 so we can do this:

[tex]m=\frac{230-220}{2009-2007}=5[/tex]

The appropiate conversion factor for this problem is:

[tex]\frac{100\operatorname{mm}}{1dm}[/tex]

Multiply the number by the given conversion factor:

[tex]6dm\cdot\frac{100\operatorname{mm}}{1dm}=600\operatorname{mm}[/tex]

The answer is 600mm.

Given:

An equation is given as

[tex]\sqrt{x+3}=x-3[/tex]

Find:

we have to solve the given equation for value of x.

Explanation:

we will solve the given equation as following

[tex]\begin{gathered} \sqrt{x+3}=x-3 \\ squaring\text{ both sides, we get} \\ x+3=x^2+9-6x \\ x^2-6x-x+9-3=0 \\ x^2-7x+6=0 \\ x^2-6x-x+6=0 \\ x(x-6)-1(x-6)=0 \\ (x-6)(x-1)=0 \\ x=6,1 \end{gathered}[/tex]

Now we will verify the solutions as

when x = 6

[tex]\begin{gathered} \sqrt{6+3}=6-3 \\ \sqrt{9}=3 \\ 3=3 \end{gathered}[/tex]

Therefore, x = 6 is the solution of given equation.

When x = 1

[tex]\begin{gathered} \sqrt{1+3}=1-3 \\ \sqrt{4}=-2 \\ 2\ne-2 \end{gathered}[/tex]

Therefore, x = 6 is the true solution of given equation.

Therefore, correct option is A, i.e. x = 6

The question can be answered directly by exploiting one of the given conditions in our question which is, the width must have a minimum of 3.2 meters.

For the LOGO 1 we have a with of 12 cm, with the scale 1 cm = 0.4 meters. Therefore our real with is;

Width 1 = 12 cm (0.4 m / 1 cm) = 4.8 m , therefore it satisfy our condition that it must have a minimum width of 3.2 meters.

while;

For the LOGO 2, we have a width of 10 cm with the scale 1 cm = 0.3 meters. Therefore our real width is;

Width 2 = 10 cm (0.3 m / 1 cm) = 3 m, making the width of our LOGO 2 not enough for our condition that the width must have a minimum of 3.2 meters.

By that alone only LOGO 1 meets the building requirements of our logo, making LOGO 1 as our answer.