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goblinko [34]
3 years ago
8

Find the second derivative of the function P(t) =(2t+4)/(t+1)

Mathematics
1 answer:
vitfil [10]3 years ago
3 0

use the quotient rule then after using it the first time use it again on your resulting equation, since you are looking for the second derivative

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is x-1 taking away an x or adding an x?? im trying to find the perimeter of area tiles, and the perimeter is x + x + x + (x-1) +
Vanyuwa [196]

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4x + 2

Step-by-step explanation:

x - 1 is adding an x and subtracting a 1 ,so

x + x + x + (x - 1) + 1 + 1 + 1

= x + x + x + x - 1 + 1 + 1 + 1 ← collect like terms

= 4x + 2

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Suppose you deposit ​$600 in a bank account with a simple interest rate of 2.5​%. You want to keep your deposit in the bank long
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120/2.5%

Step-by-step explanation:

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Find the sum and product of the roots. 3x^<br> 2 - 4x - 7 = 0
victus00 [196]
Given the quadratic function, to get the roots we factorize:
3x²-4x-7=0

3x²+3x-7x-7=0
3x(x+1)-7(x+1)=0
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6 0
3 years ago
Ana bought 324 gallons of milk last month at $3.30 per gallon and ​ 314 gallons of milk at $3.20 per gallon this month.
kipiarov [429]

Answer:

$2,074

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6 0
3 years ago
A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o
balu736 [363]

Answer:

a

P(X \ge 1) = 0.509

b

P(X  \ge 1) = 0.6807

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      P(X = x) =  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      P(X \ge 1) =  1 - P(X \le 0)

Here  

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{20 - 3} C_{4- 0}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{17} C_{4}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ 1 *  2380}{ 4845}

    P(X \le 0) =  0.491

Hence

    P(X \ge 1) =  1 - 0.491

     P(X \ge 1) = 0.509

Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

    P(X \ge 1) =  1 - P(X \le 0)

   P(X  \ge 1) =1- [  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]

Here n =  6

So

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{1  *  12376}{38760}]

    P(X  \ge 1) =1- 0.3193

    P(X  \ge 1) = 0.6807

   

5 0
3 years ago
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