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Elina [12.6K]
2 years ago
14

The volume (in m3) of water in my (large) bathtub when I pull out the plug is given by f(t)=4−t2 (t is in minutes). This formula

is only valid for the two minutes it takes my bath to drain.
(a) Find the average rate the water leaves my tub between t=1 and t=2

(b) Find the average rate the water leaves my tub between t=1 and t=1.1

(c) What would you guess is the exact rate water leaves my tub at t=1

(d) In this bit h is a very small number. Find the average rate the water leaves my tub between t=1 and t=1+h (simplify as much as possible)

(e)

What do you get if you put in h=0 in the answer to (d)?
Mathematics
1 answer:
maks197457 [2]2 years ago
8 0
The answer is EEEEEE
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An orchestra of 25 members bought tickets, at all different prices, to a concert, and the mean price paid was $82. A new musicia
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Answer:

$82.3

Step-by-step explanation:

82x25=2050

2050+90=2140

2140/26=82.3

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An certain brand of upright freezer is available in three different rated capacities: 16 ft3, 18 ft3, and 20 ft3. Let X = the ra
ruslelena [56]

Answer:

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6=349.2

V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24

The expected price paid by the next customer to buy a freezer is $466

Step-by-step explanation:

From the information given we know the probability mass function (pmf) of random variable X.

\left|\begin{array}{c|ccc}x&16&18&20\\p(x)&0.3&0.1&0.6\end{array}\right|

<em>Point a:</em>

  • The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

E(X) = $\sum_{x\in D} x\cdot p(x)

Therefore

E(X)=16\cdot 0.3+18\cdot 0.1+20\cdot 0.6=18.6

  • If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by

E[h(X)] = $\sum_{D} h(x)\cdot p(x)

So h(X) = X^2 and

E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2

  • The variance of X, denoted by V(X), is

V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2

Therefore

V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24

<em>Point b:</em>

We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:

From the rules of expected value this proposition is true:

E(aX+b)=a\cdot E(x)+b

We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore

The expected price paid by the next customer is

60\cdot E(X)-650=60\cdot 18.6-650=466

4 0
2 years ago
Margo deposited $100 into a savings account earning 4.5% simple annual interest. At the end of each year, she adds $100 to her a
nadezda [96]

Answer:

                    principal     interest              total money in account

                                                                         at the end of year

Year one       $100             $4.5                              $104.5

year two        $200          $ 9 +$4.5                         $ 213.5

year three      $300          $ 9 +$4.5 + $13.5           $ 327

Step-by-step explanation:

Simple interest for any principal is given by

I = p* r* t/100

I = interest rate accrued on principle amount

p is the amount deposited

r is the rate of interest

t is the time period of saving

_______________________________________________

For year one

p = $100

r = 4.5%

t=1

I = 100*4.5*1/100 = 4.5

_______________________________________________

For year two $100 more is added to already existing $100 in account.

p = 100 +100 = $200

r = 4.5%

t=1

I = 200*4.5*1/100 = 9

_______________________________________________

For year two $100 more is added to already existing $200 in account after two years.

p = 100 +100 +100 = $300

r = 4.5%

t=1

I = 300*4.5*1/100 = 13.5

_______________________________________________

There fore total  money in Margo account is

$300 saving deposited by her

$4.5 + $9 + $13.5 = $27 (interest accrued in three time)

Formulating the results in tabular form

                     principal     interest    total money in account

                                                            at the end of year

Year one       100               4.5                      104.5

year two        200           9 +4.5                     213.5

year three      300           9 +4.5 + 13.5           327

7 0
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3/4 = 6/x....3/4 = 6/8...notice that proportions are nothing but equivalent fractions

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