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Vera_Pavlovna [14]
3 years ago
6

Suppose that the universal set is U={1,2,3,4,5,6,7,8,9,10}. Express each of the following subsets with bit strings (of length 10

) where the ith bit (from left to right) is 1 if i is in the subset and zero otherwise.
Mathematics
1 answer:
Vladimir [108]3 years ago
5 0

Answer:

0011100000

1010010001

0111001110

Step-by-step explanation:

As the question is not complete, Here is the complete question.

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express each of these sets with bit strings where the ith bit in the string is 1 if i is in the set and 0 otherwise.

a) {3, 4, 5}

b) {1, 3, 6, 10}

c) {2, 3, 4, 7, 8, 9}.

So, we need to express a) b) and c) into bit strings.

Firstly, number of elements in the universal set represent the number of bits in the bit string.

Secondly, 1 = yes element is present in both universal set as well as in sub set.

0 = No, element is not present in sub set but present in universal set.

Hence, we have:

a) Sub set {3,4,5} = 0011100000  (As there are 3 1's which means only 3,4,5 are present in both universal set and subset.

Similarly,

b) Sub set {1, 3, 6, 10} = 1010010001

c) Sub set {2, 3, 4, 7, 8, 9} = 0111001110

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3 years ago
Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.
rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed  as:

y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]

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Answer:

The residual value is -0.75

Step-by-step explanation:

we know that  

The residual value is the observed value minus the predicted value.

RESIDUAL VALUE=[OBSERVED VALUE-PREDICTED VALUE]

where

Predicted value.--> the predicted value given the current regression equation

Observed value. --> The observed value for the dependent variable.

in this problem

we have the point (1,4)

so

The observed value is 4

<em>Find the predicted value  for x=1 </em>

y =1.5(1)+3.25=4.75

predicted value is 4.75

so

RESIDUAL VALUE=(4-4.75)=-0.75

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Isolate the variable by dividing each side by the factors that don't contain the variable

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Read 2 more answers
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