Answers
Step-by-step explanation:
the answer is in the attachment
Related Questions
If p represents a number, which expression represents "p divided by 7, increased by 8"?
Answers
Answer: p/7 + 8
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
Please help is for nowwww
Answers
Answer:
Draw a line going from 360 on the vertical line to 30 on the horizontal line.
Step-by-step explanation:
12 meters a minute. 30 minutes. 360 meters in total.
What is a sales return
Answers
Answer:
Let's say you sold some goods to a customer. If the customer is not satisfied with the goods and returns it back to you, then it is called sales return.
Help me by please this is kinda hard for my child and I forgot it
Answers
When you multiply both the numerator and denominator by the same number, you will have an equivalent to the fraction you started with because the new fraction can be reduced to the original fraction.
2/3 = 4/6 = 6/9 = 8/12 = 10/15 = 12/15 etc.
(1) Justin and Ben are scuba diving. Justin is 7 meters below the
surface of the water. Ben is 3 meters above Justin. Find Ben's
position relative to the surface of the water.
Answers
Answer:
Ben is 4 meters below the surface area.
Find the unit rate!!!
WILL GIVW BRAINLIEST!!!!
Answers
Answer:
10 to 2
Step-by-step explanation:
Please give the B
I would appreciate it :)
What is the volume of a cone with a radius of 6 inches and a height of 15 inches? (Use 3.14 for π.)
188.4 in. 3
565.2 in. 3
726.8 in. 3
1,695.6 in. 3
Thx
Answers
Answer:
1,695.6 in. 3
Step-by-step explanation:
Which statement is correct?
Thank you for anyone who answers
Answers
Answer:
I believe the correct answer is C
find the value of x in each triangle
please help me i really need help please
Answers
(x²-x+
X+5 if x 2
Suppose that h(x)= 5
if x=2
*3-1 if x>2
Which of the following is equal to 7?
I. lim hx)
X-2-
II.
lim h(x)
x->2+
III.
lim h(x)
x->2
Answers
Sorry to disappoint, but none of these answers are correct. All three limits are equal to 7.
Compute the one-sided limits:
[tex]\displaystyle\lim_{x\to2^-}h(x)=\lim_{x\to2}(x^2-x+5)=2^2-2+5=7[/tex]
[tex]\displaystyle\lim_{x\to2^+}h(x)=\lim_{x\to2}(x^3-1)=2^3-1=7[/tex]
Both the one-sided limits agree, so the two-sided limit has the same value,
[tex]\displaystyle\lim_{x\to2}h(x)=7[/tex]
City Park was receiving lots of rain. It rained 11 inches of rain in five and a half hours. What was the average rainfall for one hour?
Answers
Answer:
2
Step-by-step explanation:
11/5.5=2
5.5=5:30
Answer:
2.3
Step-by-step explanation:
What is the area of the shaded region?
Answers
Answer:
65
Step-by-step explanation:
( (23 * 13) / 2 ) - ( (13 * 13) / 2 ) = ( 299 / 2 ) - ( 169 / 2 ) = 149.5 - 84.5 = 65
Numbers are slightly unclear but I think this is right.
Area of the shaded region:
= 22 x 13 x 1/2
= 286 x 1/2
= 143
126 (x +77)
find x!!
Answers
five people were asked about the time in a week they spend in doing social work in their community they said 7,10,13,20,15 hours respectively.
which methametical concept is used in this question??
what value is depicted in this question??
Answers
Answer:
Mean time and 13
Step-by-step explanation:
Given that
here is a five people and they spend the work in 7,10,13,20,15 hours respectively
Therefore here the concept i.e. used is average time or the meantime
And, this can be determined below:
= Sum of observations ÷ number of observations
= (7 + 10 + 13 + 20 + 15) ÷ 5
= 13
MUST HELP!!!! (50 POINTS) +BRAINLEST +5 STARS +THANKS ON PROLFILE!!!!
Lines AA’, BB’, CC’ intersect at the _________
*use one of the words below to fill in the blank*
Equal 2.4 point of dilation
Scale factor perimeters (7.2,4.8)
Enlargement areas (4.8, 9.6)
Answers
[tex]\pink{▬▬▬}[/tex][tex]\red{▬▬▬}[/tex][tex]\green{▬▬▬}[/tex][tex]\blue{▬▬▬}[/tex][tex]\orange{▬▬▬}[/tex]
Equal 2.4 point of dilation
[tex]\pink{▬▬▬}[/tex][tex]\red{▬▬▬}[/tex][tex]\green{▬▬▬}[/tex][tex]\blue{▬▬▬}[/tex][tex]\orange{▬▬▬}[/tex]
Ms. Day drew a rectangle
on the board with a width
of 14 cm and a diagonal
length of 50 cm. Find the
length of this rectangle in
centimeters.
Answers
Answer:
700
Step-by-step explanation:
What is the sum of the interior angles of nonregular hexagons?
Answers
Answer:
They add up to 720 degrees
Answer:
360 I think. Hope i helped! :)
I need help its a word problemif you awnser both correctly ill give you a brainliest
Answers
Answer:
16 and 18
Step-by-step explanation:
1=5 80/5=16
1=9 2x9=18
Answer:
16 centimeters and 18 feet
Step-by-step explanation:
The scale for the fist one is 1 centimeter and that gives you 5 meters. So set up a proportion to see what 80 meters would get you. 1cm/5m=Xcm/80m. You would divide 80 by 5 to find the amount of centimeters which would be 16.
For the next one it is similar. Set up the proportion 1in/9ft=2in/Xft. You would end up with x=9times2 which would give you 18 ft.
At noon, a tank contained 20 cm of water. After several hours, it contained 18 cm of water. What is the
percent decrease of water in the tank?
The percent decrease of water in the tank is
%.
Answers
Answer:
10%Step-by-step explanation:
Percentage decrease = Difference/Original value
So,
Percentage decrease
[tex] = \frac{2}{20} \times 100 \\ = 2 \times 5 \\ = 10[/tex]
So there fore the percentage decrease is 10%
An endangered species only has 350 animals left in the wild. If the species is decreasing by 7.5% each year, how many animals are left after 2 years?
Answers
Answer:
350 X 7.5 % = 26.25
26.25 + 2 = 52.5
350 - 53 = 297 answer
Step-by-step explanation:
Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen at random from box A and placed in box B. Now, a poker chip is chosen at random from those now in box B. What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?
Answers
Answer:
0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Coin chosen from box B is red.
Event B: Blue poker chip transferred.
Probability of choosing a red coin:
7/10 of 4/9(red coin from box A)
6/10 of 5/9(blue coin from box A). So
[tex]P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444[/tex]
Blue chip transferred, red coin chosen:
6/10 of 5/9. So
[tex]P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333[/tex]
What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172[/tex]
0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red
7 points
Sue has a sticker collection with 36 red, 72 blue, and 18 red. She wants to
arrange the stickers in equal rows with only one type of sticker in each
row. Which of the following choices could Sue choose? *
18
4
12
8
Answers
What is the measure of the base in the triangle below?
B
4x + 1
3х - 8
2x + 23
11
O 45
025
30
Answers
ABC is an isosceles triangle and AB=AC.
Solving 4x+1=2x+23, we get x=11.
So, BC=3(11)-8=33-8=25
Seven students volunteered for a comparison of study guides for an advanced course in mathematics. They were randomly assigned, four to study guide A and three to study guide B. All were instructed to study independently. Following a two-day study period, all students were given an examination about the material covered by the guides, with the following results:Study Guide A scores: 68; 77; 82; 85Study Guide B scores: 53; 64; 71Perform a randomization test by listing all possible ways that these students could have been randomized to two groups. There are 35 ways. For each outcome, calculate the difference between sample averages. Finally, calculate the two-sided p-value for the observed outcome
Answers
Answer:
Step-by-step explanation:
From the given question; we can use the R software to program the combination function that generates all the combinations.
options(digits =2(
scores<- c(68,77,82,85,53,64,71)
groupA <- combn(scores,4)
groupB <- apply(groupA,2, function(x) scores[! (scores %in% x) ] )
colnames(groupA) <- colnames(groupB) <- paste("G", 1:35, sep"")
The accompanying 35 groupings (G1 to G35) contain all potential ways these understudies can be randomized under the null hypothesis
Group A
[tex]\text{G1 \ G2 \ G3 \ G4 \ G5 \ G6 \ G7 \ G8 \ G9 \ G10\ G11\ G12 \ G13 \ G14}[/tex]
[tex]\text{68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68 \ \ 68}[/tex]
[tex]\text{77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 77 \ \ 82 \ \ 82 \ \ 82 \ \ 82}[/tex]
[tex]\text{82 \ \ 82 \ \ 82 \ \ 82 \ \ 85 \ \ 85 \ \ 85 \ \ 53 \ \ 53 \ \ 64 \ \ 85 \ \ 85 \ \ 85 \ \ 53}[/tex]
[tex]\text{85\ \ 53 \ \ 64 \ \ 71 \ \ 53 \ \ 64\ \ 71\ \ 64 \ \ 71 \ \ 71 \ \ \ 53 \ \ \ 64 \ \ 71 \ \ 64}[/tex]
[tex]\text{G15 G16 G17 G18 G19 G20 G21 G22 \ G23 \ G24 \ G25 \ G26 \ G27}[/tex]
[tex]\text{68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 68 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77 \ \ \ 77}[/tex]
[tex]\text{82 \ \ \ 82 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 53 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ 85}[/tex]
[tex]\text{53 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 64 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 53}[/tex]
[tex]\text{71\ \ \ \ 71\ \ \ \ 64\ \ \ \ \ 71\ \ \ \ 71\ \ \ \ 71\ \ \ \ 53\ \ \ \ 64\ \ \ \ 71\ \ \ 64\ \ \ \ 71\ \ \ \ 71\ \ \ \ 64}[/tex]
[tex]\text{G28 G29 G30 G31 G32 G33 G34 \ G35} \\ \\ 77 \ \ \ \ 77 \ \ 77 \ \ \ \ 82\ \ \ \ 82 \ \ \ 82 \ \ \ 82 \ \ \ \ \ 85 \\ \\ 85 \ \ \ 85 \ \ \ 53 \ \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ \ 53 \ \ \ 53 \\ \\ 53 \ \ \ 64 \ \ 64 \ \ \ 53 \ \ \ \ 53 \ \ \ 64 \ \ \ \ 64\ \ \ 64 \\ \\ 71 \ \ 71 \ \ \ 71 \ \ 64 \ \ \ 71 \ \ \ \ 71 \ \ \ \ 71 \ \ \ \ 71[/tex]
Group B
[tex]\text{G1 \ G2 \ G3\ G4\ \ G5\ \ G6\ \ G7\ \ G8 \ \ G9\ \ G10\ \ G11\ \ G12\ \ G13\ G14 \ G15}[/tex]
[tex]\tet{53 \ \ 85 \ \ \ \ 85 \ \ \ \ 85\ \ \ \ 82 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 77\ \ \ \ 77\ \ \ \ 77\ \ \ \ 77\ \ \ \ 77}[/tex]
[tex]\text{64 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 85 \ \ \ 85 \ \ \ 85 \ \ \ 64 \ \ \ 53 \ \ \ 53 \ \ \ 85 \ \ \ 85}[/tex]
[tex]\text{71 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 64 \ \ \ 53 \ \ \ 71 \ \ \ 71 \ \ \ 64 \ \ \ 71 \ \ \ 64}[/tex]
[tex]\text{G16 \ G17 \ G18 \ G19 \ G20 \ G21 \ G22\ \ G23\ \ G24\ \ G25 \ \ G26 \ \ G27\ \ G28}[/tex]
[tex]\text{77\ \ \ \ 77\ \ \ \ 77\ \ \ \ \ 77\ \ \ \ \ 77\ \ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ 68\ \ \ \ \ 68\ \ \ \ \ 68\ \ \ \ \ 68}[/tex]
[tex]\text{85 \ \ \ \ 82\ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 82 \ \ \ \ 64 \ \ \ \ 53 \ \ \ \ 53 \ \ \ \ 85 \ \ \ \ 85\ \ \ \ 85 \ \ \ \ 82\ \ \ \ 82}[/tex]
[tex]\text{53\ \ \ \ 71\ \ \ \ 64\ \ \ \ 53\ \ \ \ 85\ \ \ \ 71\ \ \ \ 71\ \ \ \ 64\ \ \ \ 71\ \ \ \ 64\ \ \ \ 53\ \ \ \ 71\ \ \ \ 64}[/tex]
[tex]\text{ G29 \ G30\ G31 \ G32 \ G33 \ G34 \ G35} \\ \\ \text{68 \ \ \ 68 \ \ \ 68 \ \ \ \ 68 \ \ \ \ 68 \ \ \ 68 \ \ \ \ \ 68} \\ \\ \text{82 \ \ \ 82 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77 \ \ \ \ 77} \\ \\ \text{53 \ \ \ 85 \ \ \ \ 71 \ \ \ \ 64 \ \ \ \ 53\ \ \ \ 85 \ \ \ \ 82} \\ \\[/tex]
The accompanying data below computes the distinctions for each group:
[tex]difference <- colMeans(groupA) - colMeans(groupB)[/tex]
[tex]\text{G1 G2 G3 G4 G5 G6 G7 G8 G9 \ G10 G11 G12 G13 G14 G15} \\ \\ \text{15 -3.3 3.1 7.2 -1.6 4.8 8.9 -14 -9.8 -3.3 1.3 \ 7.8 \ 12 \ -11 \ \ -6.8}[/tex]
[tex]\text{G16 G17 G18 G19 G20 G21 G22 G23 G24 G25 G26 G27 G28} \\ \\ \text{-0.42 \ -9.2\ -5.1\ 1.3 \ -17\ \ 6.6 \ \ 13 \ 17 \ \ -5.7\ -1.6 \ \ 4.8 \ \ -3.9 \ \ 0.17}[/tex]
[tex]\text{ G29\ \ G30 \ G31 \ G32 \ \ G33 \ G34 \ \ G35} \\ \\ \text{6.6 \ \ \ -12 \ \ \ -1 \ \ \ 3.1 \ \ \ 9.5 \ \ \ -9.2 \ \ \ -7.4}[/tex]
The two-sided p-value is the extent of contrasts between test midpoints as large or bigger in supreme value than the primary group. The cat function makes the outcomes simpler to peruse.
p <- sum (aba(difference)>=difference[1])/35
cat(p)
= 0.086
The following P-Q data points (shown in Table 1) are obtained from the manufacturer's
static pressure curve for a single fan unit at 750 rpm and a density of 1.2 kg/m3. The speed is
to be increased to 850 rpm, the density changes to 1.16 kg/m3 and two fans are to be
operated in parallel.
Table 1
Quantity(
ms) 100
155
190
212
Pressure (Pa)
235
250
2000
1500
1050
500
0
Calculate the corrected single fan curve at 850 rpm and a density of 1.16 kg/m^3?
Answers
don't know
Mark as brainlist mate
because making my answer as brainlist you would not lose anything।।।।।।।।।।A desk is on sale for $217, which is 38% less than the regular price. What is the regular price?
Answers
Find the positive of x that makes the equation true.
x^2 = 225
Answers
Answer:
x = 15
Step-by-step explanation:
15 * 15 = 225
Answer:
X = 15
Step-by-step explanation:
This is because if you square 15, you get 225.
Last year, 5,200 people entered a contest to win free swimming lessons at the rec center. This year, there was 1 4% increase in the number of entries. How many entries were there this year? Enter your response in the gridded area.
Answers
2y^2 + 4y^2 + 8 + 3 helpp
Answers
Answer:
11
Step-by-step explanation:
2y + 4y = 6y = 6y + 8 + 3 Add the numbers: 8 + 3 = 11 hope this helps ;) ITS A ME LUIGI ON THE XBOX ONE WAAAAAAAAAAAAAAAAAAAAAAAAHHOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
the truth table for the formula (X → Y ) ∨ (Z → ¬X)
Answers
Answer:
[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \cline{1-4} \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \cline{1-4} \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].
[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].
Step-by-step explanation:
Let [tex]A[/tex] denote a Boolean variable.
The negation of [tex]A[/tex] ([tex]\lnot A[/tex]) is false if when [tex]\!A[/tex] is true, and true when [tex]A\![/tex] is false. In a truth table:
[tex]\begin{array}{c||c} A & \lnot A \\ \cline{1-2} \rm T & \rm F \\ \rm F & \rm T\end{array}[/tex].
[tex]\begin{array}{c||c} A & \lnot A \\ \hline \rm T & \rm F \\ \rm F & \rm T\end{array}[/tex].
Let [tex]B[/tex] denote another Boolean variable. The material implication "[tex]A[/tex] implies [tex]\!B[/tex]" ([tex]A \to B[/tex]) is true unless [tex]B\![/tex] is false when [tex]A\![/tex] is true.
[tex]\begin{array}{c|c||c} A & B & A \to B \\ \cline{1-3} \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex].
[tex]\begin{array}{c|c||c} A & B & A \to B \\ \hline \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex]
The logical or "[tex]A[/tex] or [tex]B[/tex]" is true when either [tex]A\![/tex] or [tex]B\![/tex] is true (and also when both are true.)
[tex]\begin{array}{c|c||c} A & B & A \lor B \\ \cline{1-3} \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm T \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F\end{array}[/tex]
[tex]\begin{array}{c|c||c} A & B & A \to B \\ \hline \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex].
Start by finding the value of [tex]\lnot X[/tex], [tex](X \to Y)[/tex], and [tex](Z \to \lnot X)[/tex] for each of the [tex]2^3 = 8[/tex] possible combinations of [tex]X[/tex], [tex]Y[/tex], and [tex]Z[/tex].
[tex]\begin{array}{c|c|c||c||c|c}X & Y & Z & \lnot X & (X \to Y) & (Z \to \lnot X) \\ \cline{1-6} \rm T & \rm T & \rm T & \rm F & \rm T & \rm F\\ \rm T & \rm T & \rm F & \rm F & \rm T & \rm T \\ \rm T & \rm F & \rm T & \rm F & \rm F & \rm F \\ \rm T & \rm F & \rm F & \rm F & \rm F & \rm T \\ \cline{1-6} \rm F & \rm T & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T & \rm T & \rm T \end{array}[/tex].
[tex]\begin{array}{c|c|c||c||c|c}X & Y & Z & \lnot X & (X \to Y) & (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm F & \rm T & \rm F\\ \rm T & \rm T & \rm F & \rm F & \rm T & \rm T \\ \rm T & \rm F & \rm T & \rm F & \rm F & \rm F \\ \rm T & \rm F & \rm F & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T & \rm T & \rm T \end{array}[/tex].
The value of [tex](X \to Y) \lor (Z \to \lnot X)[/tex] is true whenever either [tex](X \to Y)[/tex] or [tex](Z \to \lnot X)[/tex] is true (or both.) The combination [tex]X = \rm T[/tex], [tex]Y = \rm F[/tex], and [tex]Z = \rm T[/tex] is the only one among the eight where neither [tex](X \to Y)\![/tex] nor [tex](Z \to \lnot X)\![/tex] is true. [tex](X \to Y) \lor (Z \to \lnot X)\![/tex] would evaluate to true for all other combinations.
Hence, the truth table would be:
[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \cline{1-4} \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \cline{1-4} \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].
[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].