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3 years ago

Eurostar is a high-speed railway service connecting

Mathematics

1 answer:

zvonat [6]3 years ago

5 0

Answer:

730 trains are adequate to carry 350000 passengers.

Step-by-step explanation:

It gives us for the number of passengers, the number of seats per carriage and the number of carriages per train. To change the units from passengers to trains without changing the value, we use the multiplicative identity (that is, 1).

350000 passengers

(350000 passengers) * 1

(350000 passengers) * ((1 carriage)/(32 passengers)) * ((1 train)/(15 carriages)

[note: passengers and carriages cancel. Leaving only trains]

(350000)*(1/32)*(1/15) trains [note: I write this way to paste into MS Excel]

729.1667 trains [oh, but don’t just round this number either up or down]

729 full trains can carry 729*32*15 = 349920 passengers

730 full trains can carry 730*32*15 = 350400 passengers

Now, we can say that 730 trains are adequate to carry 350000 passengers.

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Answer:

I will say C, because tossing a coin twice has two outcomes. H is for heads while T is for heads. Tell me if I'm right.

Step-by-step explanation:

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3 years ago

In what time will the interest on $960 amount to $61.20 at 8.5% per annum?

MrRa [10]

Answer:

$525 dollars

Step-by-step explanation:

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2 years ago

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777dan777 [17]

Am not sure but Y=-x+2

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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

$b=\$300$

$y=100x+300$

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

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

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

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

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

The parametric equations x = x1 + (x2 − x1)t, y = y1 + (y2 − y1)t where 0 ≤ t ≤ 1 describe the line segment that joins the point

Ulleksa [173]

It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.

For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).

... x = 1 + t(3-1)

... y = 1 + t(4-1)

ab = {x=1+2t, y=1+3t}

For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).

... x = 3 + t(1-3)

... y = 4 + t(7-4)

bc = {x=3-2t, y=4+3t}

For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).

... x = 1 + t(1-1)

... y = 7 + t(1-7)

ca = {x=1, y=7-6t}

4 0

3 years ago