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Inga [223]
2 years ago
15

1. Writing an equation for an exponential function by

Mathematics
1 answer:
Naya [18.7K]2 years ago
5 0

Answer: 1) t(n)=0.6(2)^n

2) f(x)=10(5)^x

Step-by-step explanation:

1) Let the function that shows the thickness of the paper after n folds,

t(n) = ab^n         ---------(1)

Since, According to the question,

Initially the thickness of the paper = 0.6

That is, at n = 0, t(0) = 0.6

By equation (1),

0.6 = a(b)^0\implies 0.6 = a

Hence the function that shows the given situation,

t(n) = 0.6 b^n       -----------(2)

Again when we fold the paper the thickness of the paper will be doubled.

Thus, at n = 1, t(1) = 1.2

By equation (2),

1.2 = 0.6 b^1\implies 2 = b

Thus, the complete function is,

t(n) = 0.6 (2)^n    

2) Let the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = ab^x         ---------(1)

For f(x) = 2, x = -1

By equation (1),

2= ab^{-1}       ---------(2)

Also, For f(x) = 2/5, x = -2

Again, By equation (1),

\frac{2}{5}= a(b)^{-2}

\implies \frac{2}{5}=ab^{-1}b^{-1}=2b^{-1}

\implies \frac{2}{5}=\frac{2}{b}

\implies 2b=10

\implies b = 5

By substituting this value in equation (2),

We get, a = 10

Hence, from equation (1), the function that is passing through the points (-2, 2/5) and (-1,2),

f(x) = 10(5)^x

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Answer:

g=1.5

Step-by-step explanation:

-0.4+0.9=0.5

0.5*3 1.5

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3 years ago
An event lasts 4 hours and 20 minutes. How long did the event last in seconds?
enyata [817]
20*6=120seconds +3600second in an hour so 3600*4=14400

1210+14400=14520

14520 is the answer I hope this helps!


8 0
2 years ago
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laila [671]
I think the answer is 4/12 or 1/3
7 0
3 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
2 years ago
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