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2 years ago

College algebra online is not for me Can someone help me understand these questions

Mathematics

1 answer:

s344n2d4d5 [400]2 years ago

8 0

Answer: (1a) 20,158.7 (1b) 131,072,000

(2) F(t) = 150e²ⁿ <em>use t instead of n</em>

(3a) 8015 (3b) 8,606,040,719,360

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

$P(t) = P_oe^{kt}\\\\\bullet 2P_0=4000\\\bullet P_o=2000\\\bullet t=30\\\\\\4000=2000e^{30k}\\.\quad 2=e^{30k}\\ln(2)=30k\\\\\dfrac{ln(2)}{30}=k\qquad \longrightarrow \qquad P(t)=2000e^{\frac{ln(2)}{30}t}\\\\\\\\P(100)=2000e^{\frac{ln(2)}{30}(100)}\\.\qquad \quad =\boxed{20,158.7}\\\\\\P(6\cdot 60)=2000e^{\frac{ln(2)}{30}(8\cdot 60)}\\.\qquad \qquad = 2000 e^{16ln(2)}\\.\qquad \qquad =\boxed{131,072,000}$

***************************************************************

$F(t)=P_0e^{kt}\\\\\bullet P_0=150\\\bullet k=0.2\\\\F(t)=\boxed{150e^{0.2t}}$

****************************************************************

$P(t) = P_oe^{kt}\\\bullet P(t)=2P_o\\\bullet P_o=P_o\\\bullet t=10\\\\\underline{\text{Find k:}}\\2P_o=P_oe^{10k}\\\\2=e^{10k}\\\\ln(2)=10k\\\\\dfrac{ln(2)}{10}=k\\\\\\\bullet P(90)=50,000\\\bullet P_o=P_o\\\bullet t=9\\\\\underline{\text{Find}\ P_o:}\\\\50,000=P_oe^{\frac{ln(2)}{10}90}\\\\\dfrac{50,000}{e^{9ln(2)}}=P_o\\\\\boxed{8015}=P_o\qquad \longrightarrow \qquad P(t)=8015e^{\frac{ln(2)}{10}t}$

$P(5\cdot 60)=8015e^{\frac{ln(2)}{10}(5\cdot 60)}\\.\qquad \qquad = 8015 e^{30ln(2)}\\.\qquad \qquad =\boxed{8,606,040,719,360}$

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Determine algebraically whether the function is even, odd, or neither even nor odd.

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Answer:

odd

Step-by-step explanation:

Just so you know there are shortcuts for determining if a polynomial function is even or odd. You just to make sure you use that x=x^1 and if you have a constant, write it as constant*x^0 (since x^0=1)

THEN!

If all of your exponents are odd then the function is odd

If all of your exponents are even then the function is even

Now you have -4x^3+4x^1

3 and 1 are odd it is an odd function

This a short cut not the legit algebra way

let me show you that now:

For it to be even you have f(-x)=f(x)

For it be odd you have f(-x)=-f(x)

If you don't have either of those cases you say it is neither

So let's check

plug in -x -4(-x)^3+4(-x)=-4*-x^3+-4x=-4x^3+-4x

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2 years ago

. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$P(U |D ) = 0.198$

$P(O\ n \ B) = 0.188$

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

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$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

$P(O \ n \ B) = \frac{117}{623}$

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Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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Answer:

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Step-by-step explanation:

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