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Anit [1.1K]
3 years ago
10

Consider the following functions.G(x) = 4x2; f(x) = 8x(a)

Mathematics
1 answer:
noname [10]3 years ago
8 0

Answer:

(a) B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

(b) Every function of the form 4x^2+C is an antiderivative of 8x

Step-by-step explanation:

A function <em>F </em>is an antiderivative of the function <em>f</em> if

F'(x)=f(x)

for all x in the domain of <em>f.</em>

(a) If f(x) = 8x, then G(x)=4x^2 is an antiderivative of <em>f </em>because

G'(x)=8x=f(x)

Therefore, G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

Let F be an antiderivative of f. Then, for each constant C, the function F(x) + C is also an antiderivative of <em>f</em>.

(b) Because

\frac{d}{dx}(4x^2)=8x

then G(x)=4x^2 is an antiderivative of f(x) = 8x. Therefore, every antiderivative of 8x is of the form 4x^2+C for some constant C, and every function of the form 4x^2+C is an antiderivative of 8x.

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A puppy and kitten are 180 meters apart when they see each other. The puppy can run at a speed of 25 meters per second, while th
Gnom [1K]

It seems it should be the other way... . How soon will the puppy catch up with the kitten.

The way you have it the kitten will never catch the pup. There is already 180m between them so, the distance will only increase because the puppy runs faster than the cat.

If the puppy runs after the cat, will eventually catch up with it and can be calculated. Not the other way around.

If the pup runs after the cat, the distance between them will decrease and the pup will catch the cat.

RT = D kitten

RT = D+180 puppy

20T = D

25T = D + 180

Substitution

25T = 20T +180

5T = 180

T = 36s

7 0
3 years ago
Write a recursive formula for
algol13

9514 1404 393

Answer:

  • a[1] = 3
  • a[n] = a[n-1] -7

Step-by-step explanation:

A recursive formula consists of two parts:

  • initialization (rule for the first term(s))
  • rule for the next term

When we look at the differences between terms in the sequence 3, -4, -11, ..., we find that they are constant at -7. That is each term can be found from the previous one by subtracting 7. This is our recursive rule. The first term is obviously 3, so the recursive formula is ...

  a[1] = 3

  a[n] = a[n-1] -7

5 0
3 years ago
Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random
yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is \mu*n and the standard deviation is s = \sigma \sqrt{n}

In this problem, we have that:

\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

Z = \frac{4000 - 3500}{160}

Z = 3.13

Z = 3.13 has a pvalue of 0.9991

X = 3000

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

Z = \frac{3000 - 3500}{160}

Z = -3.13

Z = -3.13 has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0
3 years ago
Why 10^33/2=5*10^32? Explain, pls. :(
german

Answer:

see below

Step-by-step explanation:

10^33/2

Rewriting the numerator as 10 * 10 ^ 32

10 * 10 ^32

--------------------

2

10/2 = 5

5 * 10 ^32

Therefore

10^33/2=5*10^32

4 0
2 years ago
Noriko multiplies 13 × 45.<br> Which of the following is NOT a partial product
LiRa [457]

Answer:

585

Step-by-step explanation:

7 0
2 years ago
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