Answer: 28.26
Step-by-step explanation:
the truth table for the formula (X → Y ) ∨ (Z → ¬X)
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].
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?
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
(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
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]
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
a dog trainer has 104ft of fencing that will be used to create a retangular work area for dogs. If the trainer wants to enclosed an area of 532ft what will be the dimensions of the work area?
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.
Helpppp please due in 29 minutes!!!!!
Answer:
d is the answer ask me more question
126 (x +77)
find x!!
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?
Answer:
2
Step-by-step explanation:
11/5.5=2
5.5=5:30
Answer:
2.3
Step-by-step explanation:
Find the unit rate!!!
WILL GIVW BRAINLIEST!!!!
Answer:
10 to 2
Step-by-step explanation:
Please give the B
I would appreciate it :)
(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.
Answer:
Ben is 4 meters below the surface area.
I need help its a word problemif you awnser both correctly ill give you a brainliest
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.
2y^2 + 4y^2 + 8 + 3 helpp
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
Which expression is equivalently to x^2 + 2x + 2?
Answer:
(X-1+I)(x-1-I)
Step-by-step explanation:
Expression (x + 1 - i)(x + 1 + i) is equivalent to x² + 2x +2.
Option (B) is correct option.
What is an algebraic expression?Variables, integers, and a variety of mathematical procedures make up an algebraic expression.
To find the expression equivalent to x² + 2x +2,
Solve with the help of options.
(A)
(x - 1 + i)(x - 1 - i)
Solve by using formula (a + b)(a - b) = a² - b²
(x - 1 + i)(x - 1 - i) = (x -1)² - i² = x² + 1 - 2x +1 = x² - 2x + 2
(B)
(x + 1 - i)(x + 1 - i)
Solve by using formula (a + b)(a - b) = a² - b²
(x + 1 - i)(x + 1 + i) = (x + 1)² - i² = x² + 1 + 2x +1 = x² + 2x + 2
Expression in option (B) is equivalent to x² + 2x + 2.
To know more about Algebraic expression on:
https://brainly.com/question/19245500
#SPJ2
Please help is for nowwww
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 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
Answer:
1,695.6 in. 3
Step-by-step explanation:
Jaclyn plays a game of chance. Each play is independent. Each time that she plays, the probability that she wins is .3. She plays 16 times. What is the probability that she will win exactly 2 times
Answer:
8
Step-by-step explanation:
The probability that Jaclyn wins exactly 2 times is 0.073
What is binomial probability?Binomial probability is the probability of exactly x successes on n repeated trials in an experiment with two possible outcomes. If the probability of success on an individual trial is p, then the binomial probability is ⁿCₓ⋅pˣ⋅(1−p)ⁿ⁻ˣ. Here ⁿCₓ indicates the number of different combinations of x objects selected from a set of n objects.
Given here:
She plays 16 times and the probability of her winning is 0.3
Thus applying the formula we get
= ¹⁶C₂× 0.3²× 0.7¹⁴
= 120×0.09×0.7¹⁴
=0.073
Hence, The probability that Jaclyn wins exactly 2 times is 0.073
Learn more about binomial probability here:
https://brainly.com/question/29350029
#SPJ5
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.
Answer:
700
Step-by-step explanation:
If p represents a number, which expression represents "p divided by 7, increased by 8"?
Answer: p/7 + 8
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
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??
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)
[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]
Help me by please this is kinda hard for my child and I forgot it
A desk is on sale for $217, which is 38% less than the regular price. What is the regular price?
What is a sales return
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.
find the value of x in each triangle
please help me i really need help please
What is the area of the shaded region?
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.
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?
don't know
Mark as brainlist mate
because making my answer as brainlist you would not lose anything।।।।।।।।।।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
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
What is the measure of the base in the triangle below?
B
4x + 1
3х - 8
2x + 23
11
O 45
025
30
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
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?
Answer:
350 X 7.5 % = 26.25
26.25 + 2 = 52.5
350 - 53 = 297 answer
Step-by-step explanation: