What is the domain of the function shown in thr table

Decrypt the word DKA if it was encrypted using an alphabetic Caesar shift cipher that starts with shift 5 (mapping A to F), and

pochemuha

Answer:

YET

Step-by-step explanation:

TO DECRYPT DKA USING THE CAESAR SHIFT CIPHER STARTING WITH 5 SHIFTS.

TO DO THIS DECRYPTION, WE'RE GOING TO LOOP THE ALPHABETS.

D = LOOPING THIS LETTER 5 TIMES, WE HAVE C - B - A - Z - Y.

SO, D IS DECRYPTED AS Y

GOING FURTHER, THE ENCRYPTION SHIFTS BY AN ADDITIONAL SPACE. THIS MEANS, IF THE FIRST "Y" WAS ENCRYPTED IN 5 SPACES TO GET D, AND AN ADDITIONAL SPACE WAS ADDED TO THE NEXT ALPHABET, IT BECOMES 6.

K IS THEN DECRYPTED AS

K = J - I - H - G - F - E

K IS DECRPYTED AS E

GOING FURTHER, THE NEXT WAS ENCRYPTED WITH AN ADDITIONAL SPACE EVEN MORE, SO, A IS DECRPYTED AS

A = Z - Y - X - W - V - U - T.

THUS, DKA IS DECRYPTED AS YET

5 0

3 years ago

The inequality x + 2y ≥ 3 is satisfied by point (1, 1). True False

andre [41]

In the given statement above, in this case, the answer would be TRUE. It is true that the inequality x + 2y ≥ 3 is satisfied by point (1, 1). In order to prove this, we just have to plug in the values. 1 + 2(1) ≥ 3
So the result is 1 + 2 ≥ 3. 3 ≥ 3, which makes it true, because it states that it is "more than or equal to", therefore, our answer is true. Hope this answer helps.

7 0

3 years ago

Read 2 more answers

14 yd

son4ous [18]

Answer:

The volume is 196 cubic yards

Step-by-step explanation:

Given

$Base\ Area = 7yd * 2yd$

$Height = 14yd$

Required

The volume

The volume is given as:

$Volume = Base\ Area * Height$

Where:

$Base\ Area = 7yd * 2yd$

$Height = 14yd$

So, the equation becomes

$Volume = 7yd * 2yd * 14yd$

$Volume = 196yd^3$

6 0

3 years ago

What conclusion can be made based on this multiplication problem?

Elena L [17]

Answer:

I believe it is twenty-one is 7 times the size of 3

Step-by-step explanation:

4 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago