Freeeeee point because i said so ughh

Yes, you are correct. Suppose the relationship is p=aq. Put p=10 and q=2 in, we can get p=5q. So when q=20, p=20*5=100. So the answer is D. Be confident with yourself~

5 0

3 years ago

At a certain electronics company, the daily output Q is related to the number of people A on the assembly line by Q=600+√(A+41).

Ber [7]

Answer:

$A=400\ people$

Step-by-step explanation:

we have

$Q=600+\sqrt{A+41}$

where

Q -----> is the daily output

A -----> is the number of people

For Q=621

Find the value of A

substitute and solve for A

$621=600+\sqrt{A+41}$

Subtract 600 both sides

$621-600=\sqrt{A+41}$

$21=\sqrt{A+41}$

squared both sides

$21^{2} ={A+41}$

$441={A+41}$

Subtract 41 both sides

$441-41=A$

Rewrite

$A=400\ people$

3 0

2 years ago

List the coordinates of FOUR vertices that create the feasible region on the graph. Submit your answer in the form of FOUR order

hoa [83]

Answer:

The coordinates of the vertices are;

(200, 200), (300, 200), (300, 0), (0, 500)

Step-by-step explanation:

The vertices are the corners of a polygon. It is the point where an angle is formed by the intersection of two lines

The feasible region is the solution space of the points of the variables that meet the specification of the problem set by means of a constant or the definition of inequalities or equations

From the graph of the function of inequalities, we have that the four vertices are the four points where the lines bounding the area of the feasible region of the inequalities meet

The coordinates of the vertices are (200, 200), (300, 200), (300, 0), (0, 500).

4 0

2 years ago

John, Sally, and Natalie would all like to save some money. John decides that it

brilliants [131]

Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) $\$2,700$

Part 5) Is a exponential growth function

Part 6) $A=6,000(1.07)^{t}$

Part 7) $\$11,802.91$

Part 8) $\$6,869.40$

Part 9) Is a exponential growth function

Part 10) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 11) $\$13,591.41$

Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

y=mx+b

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

so

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{2}=\$6,869.40$

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

$A=5,000(e)^{0.10*2}=\$6,107.01$

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

3 0

3 years ago

Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights

Fed [463]

Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

Given that, A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171, Z = (4171 - 3311)/860 = 1

P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.

Next, multiply that by the sample size of 732.

Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171

For part b, use the same method except x is now 1591.

Z = (1581 - 3311)/860 = -2

P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

5 0

2 years ago