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2 years ago

HELPP ASAP NEEDED MATH

Mathematics

2 answers:

trapecia [35]2 years ago

6 0

Answer:

992

Step-by-step explanation:

Divide 1000 by 26.

The answer is 38 and some left over. We don't care what the leftover is because it is nearly 0.5 and that means 13 people were left over.

Take the integer value (38) and multiply it by 26. You get 988.

You want there to be 4 left over. 4 + 988 = 992. That's one way of doing the problem.

Umnica [9.8K]2 years ago

3 0

Answer: 940 people is the maximum

==========================================

Explanation:

n = 1 means 20n = 20*1 = 20 members total, which is 4 short of making a row of 26 (so the remainder is 20)

n = 2 means 20n = 20*2 = 40 members. 40/26 = 1 remainder 14. We could make one full row with 14 members left over.

20n < 1000 solves to n < 50 after dividing both sides by 20. This means n is smaller than 50. It has to be larger than 0 or else 20n would be 0 or negative. We can write 0 < n < 50 to specify the range for n.

Since n < 50, this means the largest n can get is n = 49.

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

If n = 49, then 20n = 20*49 = 980

980/26 = 37 remainder 18, which isn't the remainder we want (of 4).

If n = 48, then 20n = 20*48 = 960

960/26 = 36 remainder 24, again not the remainder we want

If n = 47, then 20n = 20*47 = 940

940/26 = 36 remainder 4

If we have 940 members, then we can form 36 rows (26 in each row). So 26*36 = 936 people so far. Then we have 940-936 = 4 people left over.

There are other solutions as well such as n = 34, n = 21, and n = 8. Though n = 47 is the largest n possible to fit all the conditions required.

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2 years ago

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We flip a fair coin five times. For every heads you pay me $1 and for every tails I pay you $1. Let X denote the my net winning

TEA [102]

Answer:

The probability mass of X is 0.03

Step-by-step explanation:

If we set the winning requirement of your heads and my tails then the occurring possibility of both is 1/2 or 0.5.

Hence let us make a graph and use the figures to calculate the all the probabilities of you getting a heads.

Where X represents the number of dollars won during the flip of the coin, probability of heads represent the chances of occurrence of the value and of winning the dollars.

The probability of winning start to drop as the winning amount increases.

X 0 1 2 3 4 5

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2 years ago

Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form $p(t)=at^2$ , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$ , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$ , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

$p_1 (t) = at^2, p_2(t)=bt^2$ .

We must check that p_1+p_2(t) is in W1.

Note that

$p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2$

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say $p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)$ . Then

$p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)$

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 $p(t) = at^2$ and a real scalar b. Then

$bp(t) = b(at^2) = (ba)t^2)$ .

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 $p(t) = a(t^2+t)$ and a real scalar b. Then

$bp(t) = b(a(t^2+t)) = (ba)(t^2+t)$ .

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.

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2 years ago

The caller times at a customer service center has an exponential distribution with an average of 22 seconds. Find the probabilit

jenyasd209 [6]

Answer:

The probability that a randomly selected call time will be less than 30 seconds is 0.7443.

Step-by-step explanation:

We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.

Let X = caller times at a customer service center

The probability distribution (pdf) of the exponential distribution is given by;

$f(x) = \lambda e^{-\lambda x} ; x > 0$

Here, $\lambda$ = exponential parameter

Now, the mean of the exponential distribution is given by;

Mean = $\frac{1}{\lambda}$

So, $22=\frac{1}{\lambda}$ ⇒ $\lambda=\frac{1}{22}$

SO, X ~ Exp( $\lambda=\frac{1}{22}$ )

To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;

$P(X\leq x) = 1 - e^{-\lambda x}$ ; x > 0

Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)

P(X < 30) = $1 - e^{-\frac{1}{22} \times 30}$

= 1 - 0.2557

= 0.7443

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2 years ago

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Dafna1 [17]

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