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Igoryamba
3 years ago
15

7. Simplify this what’s the answer?

Mathematics
1 answer:
trasher [3.6K]3 years ago
3 0

first the square root of 4 is 2 so we can sub that in.

now we have 2(2*2)^-2.

2 * 2 of course is 4

with 2(4)^-2, keep in mind that negative exponents just mean to "flip" the number and turn the exponents positive.

so 4^-2 is the same as 1/4^2 or 1/16.

finally 1/16 * 2 = 2/16. 2/16 simplified is 1/8.

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16400*0.0825= 1353

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Answer:

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Step-by-step explanation:

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3 years ago
Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are
Svetlanka [38]

Answer:

(a) Number of sets B given that

  • A⊆B⊆C: 2¹⁰.  (That is: A is a subset of B, B is a subset of C. B might be equal to C)
  • A⊂B⊂C: 2¹⁰ - 2.  (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let x_1, x_2, \cdots, x_{20} denote the 20 elements of set X.

Let x_1, x_2, \cdots, x_{10} denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain x_1, x_2, \cdots, x_{10}.

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from x_1, x_2, \cdots, x_{20}.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 2^{10} -2 = 1024 - 2 = 1022 possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's 2^{20} = 1048576 possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 2^{20} - 2^{10} possibilities left.

5 0
2 years ago
Give the largest interval over which the general solution is defined. (Think about the implications of any singular points. Ente
Leto [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   No singular point due to the exponent in the solution

    The interval is   -\infty

b

   NONE

Step-by-step explanation:

From the question we are told that

     \frac{dy}{dx} =  9y

The generally solution is mathematically represented as

         \frac{dy }{dx}  =  9y

=>       \frac{dy}{y}  =  9dx

integrating both sides  

         \int\limits  {\frac{ dy}{y} } \,  = \int\limits  {9} \, dx

  =>   lny = 9x + c

 =>   y =  e^{9x +c }

 =>    y =  e^{9x} e^{c}

Here e^c  =  C

=>     y = C  e^{9x}

From the above equation we see that the domain for x has no singular point the interval is

       -\infty

Also there is no transient term in the general solution obtained because as  x \to \infty there no case where y \to 0

7 0
3 years ago
Match its group with its percentage or mean Using
krok68 [10]

(1)

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<u>Mean</u> = (Sum of observations)/(Number of observations)

= (12 + 5 + 3 + 5 + 8 + 2 + 10 + 9 + 4 + 4)/(10)

= 62/10

=<em> 6.2</em>

(2)

Mean length of adult fish in the sample - \boxed{\sf{\red{8.75}}}

<u>Mean</u> = (Sum of observations)/(Number of observations)

= (12 + 5 + 8 + 10)/(4)

= 35/4

=<em> 8.75</em>

(3)

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<u>Mean</u> = (Sum of observations)/(Number of observations)

= (5 + 3 + 2 + 9 + 4 + 4)/(6)

= 27/6

<em>= 4.5</em>

(4)

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<u>Percentage</u> = (No. of adult fishes)/(Total no. of fishes) × 100

% = (4/10) × 100

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(5)

Percentage of sample that were juvenile fish - \boxed{\sf{\red{60}}}

<u>Percentage</u> = (No. of juvenile fishes)/(Total no. of fishes) × 100

% = (6/10) × 100

<em>% = </em><em>6</em><em>0</em>

(6)

Percentage of sample that were juveniles over 8 inches long - \boxed{\sf{\red{10}}}

<u>Percentage</u> = (No. of juveniles over 8 inches)/(Total no. of fishes) × 100

% = (1/10) × 100

<em>% = </em><em>1</em><em>0</em>

7 0
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