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Nadusha1986 [10]
4 years ago
11

What is the sum of all the integers between the square root of 19 and the square root of 77?

Mathematics
1 answer:
Artist 52 [7]4 years ago
4 0
Since we are only dealing with integers, we do not need to worry about decimals. Therefore, finding the exact values of sqrt(19) and sqrt(77) is not important. We can just estimate.

19 is between 16 and 25 (which are 4^2 and 5^2), so the square root of 18 is some number between 4 and 5.

77 is between 64 and 81 (which are 8^2 and 9^2), so the square root of 77 is some number between 8 and 9.

So the integers (whole numbers) that are between our two sqaure roots are 5, 6, 7, and 8.

Add them up and we have our answer.

5+6+7+8 = 26




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Answer:

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And we can find this probability using the complement rule:

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And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X \sim N(440,20)  

Where \mu=440 and \sigma=20

We are interested on this probability

P(X>440)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>400)=P(\frac{X-\mu}{\sigma}>\frac{400-\mu}{\sigma})=P(Z>\frac{400-440}{20})=P(z>-2)

And we can find this probability using the complement rule:

P(z>-2)=1-P(z

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(z>-2)=1-P(z

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Read 2 more answers
For each, list three elements and then show it is a vector space.
Butoxors [25]

Answer:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

2+3x \in P_1

4.5+\sqrt2 x \in P_1

\log5+78x \in P_1

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

4+2x \in P_2

13+6.5x \in P_2

10.5+5.25x \in P_2

Step-by-step explanation:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

  • 2+3x \in P_1
  • 4.5+\sqrt2 x \in P_1
  • \log5+78x \in P_1

A vector space is any set whose elements hold the following axioms for any \vec{u}, \vec{v} and \vec{w} and for any scalar a and b:

  1. (\vec{u} + \vec{v} )+\vec{w} = \vec{u} +( \vec{v} +\vec{w})
  2. There is the <em>zero element </em>such that: \vec{0} + \vec{u} = \vec{u} + \vec{0}
  3. For all element \vec{u}of the set, there is an element -\vec{u} such that: -\vec{u} + \vec{u} = \vec{u} + (-\vec{u}) = \vec{0}
  4. \vec{u} + \vec{v} = \vec{v} + \vec{u}
  5. a(b\vec{v}) = (ab)\vec{v}
  6. 1\vec{u} = \vec{u}
  7. a(\vec{u} + \vec{v} ) = a\vec{u} + a\vec{v}
  8. (a+b)\vec{v} = a\vec{v}+b\vec{v}

Let's proof each of them for the first set. For the proof, I will define the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g.

  1. (a_0+a_1x + b_0+b_1x)+c_0+c_1x = a_0+a_1x +( b_0+b_1x+c_0+c_1x)\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we obtain \boxed{\alpha_0+\alpha_1x= \alpha_0+\alpha_1x} which is another polynomial that belongs to P_1
  2. A null polynomial is define as the one with all it coefficient being 0, therefore: \boxed{0 + a_0+a_1x = a_0+a_1x + 0 = a_0+a_1x}
  3. Defining the inverse element in the addition as -a_0-a_1x, then -a_0-a_1x + a_0 + a_1x = a_0+a_1x + (-a_0-a_1x)\\\boxed{(-a_0+a_0)+(-a_1+a_1)x = (a_0-a_0)+(a_1-a_1)x = 0}
  4. (a_0+a_1x) +( b_0+b_1x) =( b_0+b_1x) +( a_0+a_1x)\\(a_0+b_0)+(a_1+b_1)x = (b_0+a_0)+(b_1+a_1)x\\\boxed{(a_0+b_0)+(a_1+b_1)x = (a_0+b_0)+(a_1+b_1)x}
  5. a[b(a_0+a_1x)] = ab (a_0+a_1x)\\a[ba_0+ba_1x] = aba_0+aba_1x\\\boxed{aba_0+aba_1x = aba_0+aba_1x}
  6. \boxed{1 \cdot (a_0+a_1x) = a_0+a_1x}
  7. \boxed{a[(a_0+a_1x)+(b_0+b_1x)] = a(a_0+a_1x) + a(b_0+b_1x)}
  8. (a+b)(a_0+a_1x)=aa_0+aa_1x+ba_0+ab_1x\\\boxed{(a+b)(a_0+a_1x)= a(a_0+a_1x) + b (a_0+a_1x)}

With this, we proof the set P_1 is a vector space with the usual polynomial addition and scalar multiplication operations.

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

  • 4+2x \in P_2
  • 13+6.5x \in P_2
  • 10.5+5.25x \in P_2

Let's proof each of axioms for this set. For the proof, I will define again the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g. Again the relation a_1 = \frac{a_0}{2} between the coefficients holds

  1. [(a_0+a_1x) +( b_0+b_1x)]+(c_0+c_1x) = (a_0+a_1x) +[( b_0+b_1x)+(c_0+c_1x)]\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and considering the coefficient relation and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we have (a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x\\(a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x = (a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x\\\boxed{\alpha_0 + \alpha1x = \alpha_0 + \alpha1x} which is another element of the set since it is a degree one polynomial whose coefficient follow the given relation.

The proof of the other axioms can be done using the same logic as in (a) and checking that the relation between the coefficients is always the same.

6 0
4 years ago
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