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3 years ago

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The video states that 78% of people carry less than $50 in cash; 40% carry less than $20 in cash, and 9% carry no cash. Suppose

you have a random sample of 100 people and you check how much cash each person is carrying. Assuming that the percentages stated in the video are correct for all adult Americans, discuss how it could occur that 12 people in your sample are carrying no cash while only 75 people in your sample are carrying less than $50 in cash.

1 answer:

Inessa05 [86]3 years ago

6 0

Answer:

The probability that 12 people in your sample are carrying no cash is 0.0712

Step-by-step explanation:

n = 100

p(no cash) = 0.09

x = 12

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 12) = 0.074.

The probability that 12 people in your sample are carrying no cash is 0.074.

n = 100

p(less than 50) = 0.78

x = 75

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 75) = 0.0712

The probability that 12 people in your sample are carrying no cash is 0.0712

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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

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