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3 years ago

What is the GCF of and 55 and 80 *

Mathematics

1 answer:

Zarrin [17]3 years ago

3 0

The Greatest Common Factor of 55 and 80 is 5.

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Fundamental theorem of calculus <img src="https://tex.z-dn.net/?f=g%28s%29%3D%5Cint%5Climits%5Es_6%20%7B%28t-t%5E4%29%5E6%7D

mr_godi [17]

Answer:

$\displaystyle g'(s) = (s-s^4)^6$

Step-by-step explanation:

The Fundamental Theorem of Calculus states that:
$\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)$

Where a is some constant.

We can let:
$\displaystyle g(t) = (t-t^4)^6$

By substitution:

$\displaystyle g(s) = \int_6^s g(t)\, dt$

Taking the derivative of both sides results in:
$\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]$

Hence, by the Fundamental Theorem:

$\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\ & = (s-s^4)^6\end{aligned}$

3 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago

You are to take a multiple-choice exam consisting of 64 questions with 5 possible responses to each question. Suppose that you h

damaskus [11]

Answer:

a) Expected score on the exam is 12.8.

b) Variance 10.24, Standard deviation 3.2

Step-by-step explanation:

For each question, there are only two possible outcomes. Either you guesses the answer correctly, or you does not. The probability of guessing the answer of a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

The standard deviation of the binomial distribution is:

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

64 questions.

So $n = 64$

5 possible answers, one correctly, chosen at random:

So $p = \frac{1}{5} = 0.2$

(a) What is your expected score on the exam?

$E(X) = np = 64*0.2 = 12.8$

(b) Compute the variance and standard deviation of x. Variance =Standard deviation

$V(X) = np(1-p) = 64*0.2*0.8 = 10.24$

Variance 10.24

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{64*0.2*0.8} = 3.2$

Standard deviation 3.2

5 0

3 years ago

Let f be continuous on [0, a] and differentiable on (0, a), Prove that if f(a)=0 then there is at least one value of x in (0, a)

asambeis [7]

Answer:

See picture attached

Step-by-step explanation:

6 0

3 years ago

A circular swimming pool has a diameter of the 26 feet.

kherson [118]

Hey there! :)

The formula for finding area of circles is: π · r²

Remember that radius is half the diameter. To find radius when the diameter is given, divide by 2

d = 26

26 ÷ 2 = 13
_____________________________________________________________

π · r²

r = 13
13² = 13 × 13 = 169

169 × 3.14 = 530.66 = 530.7

The area of the swimming pool is 530.7 ft²

Hope this helps :)

7 0

3 years ago

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