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alexira [117]
3 years ago
14

Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)

= te^11-6t5) g(p) = pIn(2p+3)6) z= (te^6t + e^5t)^77) w= 2y+y^2 / 7+y
Mathematics
1 answer:
borishaifa [10]3 years ago
7 0

Answer:

Since,

\frac{d}{dx}x^n = nx^{n-1}

\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))

\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}

1) y = 8x e^x

Differentiating with respect to x,

\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)

2) y = 5x e^{x^4}

Differentiating w. r. t x,

\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)

3) y = x^8 + \frac{5}{x^4}

Differentiating w. r. t. x,

\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}

4) f(t) = te^{11}-6t^5

Differentiating w. r. t. t,

f'(t) = e^{11} - 30t^4

5) g(p) = p\ln(2p+3)

Differentiating w. r. t. p,

g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)

6) z = (te^{6t}+e^{5t})^7

Differentiating w. r. t. t,

\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})

7) w =\frac{2y + y^2}{7+y}

Differentiating w. r. t. y,

\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}

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A college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-st
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Answer:

0 < x ≤ 100 and 0 < y ≤ 300

Step-by-step explanation:

THIS IS THE COMPLETE QUESTION BELOW;

college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-state students as out-of-state, and they only have space to accept 100 out-of-state students. Let x = the number of out-of-state students and y = the number in-state students. Write the constraints to represent the incoming students at the college.

0 < x ≤ 100 and 0 < y ≤ 300

x > 0 and y > 0

0 < x ≤ 100 and y > 300

0 < x and y < 100

SOLUTION

they only have space to accept 100 out-of-state students,which means that the Maximum number of out-of-state students that can be accepted is 100

Then x= 100(Maximum number of out-of-state students that can be accepted)

They plan to accept three times as many in-state students as out-of-state which means that

Y = 3x(Maximum number of in-state students)

Then we can deduced that the numbee out-of-state students that can be accepted can lyes between the range of 0 and 100 which means from interval 0 to 100

Which can be written as 0 < x ≤ 100

But we need to know the interval for the Maximum number of in-state students(Y), to do that we need to multiply the equation above by 3 since Y = 3x

0 < x ≤ 100

3× 0 = 0

3× X = X

3× 100= 300

Then 0 < 3x≤ 300

But we know that Y = 3x then substitute into last equation

We have

0 < y ≤ 300

ThenBthe constraints to represent the incoming students at the college is

0 < x ≤ 100 and 0 < y ≤ 300

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2 years ago
According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
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Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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