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Vinil7 [7]
3 years ago
10

Tn+1=35-2Tn, T1=5 what is T20?

Mathematics
1 answer:
lbvjy [14]3 years ago
4 0

Answer:  3,495,265

<u>Step-by-step explanation:</u>

T_{n+1}=35-2T_n\quad and\quad T_1=5, then\\T_{1+1}=35-2(T_1)\implies T_2=35-2(5)\implies T_2=25\\T_{2+1}=35-2(T_2)\implies T_3=35-2(25)\implies T_3=-15\\T_{3+1}=35-2(T_3)\implies T_4=35-2(-15)\implies T_4=65\\T_{4+1}=35-2(T_4)\implies T_5=35-2(65)\quad \implies T_5=-95\\T_{5+1}=35-2(T_5)\implies T_6=35-2(-95)\implies T_6=225\\.\qquad \qquad \qquad \qquad \qquad \qquad \downarrow\\.\qquad \qquad \qquad \qquad \qquad \qquad \downarrow\\.\qquad \qquad \qquad \qquad \qquad \qquad \downarrow

T_{18+1}=35-2(T_{18})\implies T_{19}=35-2(873,825)\implies T_{19}=-1,747,615\\T_{19+1}=35-2(T_{19})\implies T_{20}=35-2(-1,747,615)\implies T_{20}=3,495,265

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The mean and range for class A are 84.4 and 20 while the mean and range for class B are 83.6 and 8. Class B is more consistant then class A.

<h3>How to calculate the mean?</h3>

We can see from the data that the consistency of class B is more becuase the repeatation of the same score is more in class B then class A  and the range of class B is also less then the class A.

It should be noted that mean simply means average. It's the addition of the numbers divided by the total numbers.

In this case, the computation will be:

Class A: Mean = 422/5= 84.4.

Range: 94-74 = 20

Class B: Mean = 418/5 = 83.6.

Range= 88-80 = 8

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I believe this is the answer you are looking for:

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3 years ago
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3 years ago
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3 years ago
Solve for y 16y^2-25=0
Pepsi [2]

Answer:

\large\boxed{x=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

Step-by-step explanation:

16y^2-25=0\\\\METHOD\ 1:\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\16=4^2\ \text{and}\ 25=5^2\ \text{therefore we have}\\\\4^2y^2-5^2=0\\\\(4y)^2-5^2=0\\\\(4y-5)(4y+5)+0\iff4y-5=0\ \vee\ 4y+5=0\\\\4y-5=0\qquad\text{add 5 to both sides}\\4y=5\qquad\text{divide both sides by 4}\\\boxed{y=\dfrac{5}{4}}\\\\4y+5=0\qquad\text{subtract 5 from both sides}\\4y=-5\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{5}{4}}

METHOD\ 2:\\\\16y^2-25=0\qquad\text{add 25 to both sides}\\\\16y^2=25\qquad\text{divide both sides by 16}\\\\y^2=\dfrac{25}{16}\to y=\pm\sqrt{\dfrac{25}{26}}\\\\y=-\dfrac{\sqrt{25}}{\sqrt{16}}\ \vee\ x=\dfrac{\sqrt{25}}{\sqrt{16}}\\\\\boxed{y=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

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3 years ago
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