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Alexxandr [17]
2 years ago
8

The table of values shows the number of touchdown passes that a quarterback threw in each of his first five football games of th

e season.
Game vs. Touchdown Passes
Game Touchdown passes
1............................3
2............................ 1
3............................. 3
4........................... 2
5........................... 3

Which would be a point on the scatterplot that represents touchdown passes based on these games played?
(1, 2)
(2, 4)
(3, 1)
(5, 3)
Mathematics
2 answers:
tatuchka [14]2 years ago
7 0
The point on the scatter plot is (5,3) which represents the number of touchdown passes based on these games played.
kvv77 [185]2 years ago
3 0
If x is the game number, and y is the amount of touchdown passes for that specific game number, then (5,3) represents the idea that the quarter back threw y = 3 touchdown passes for game number x = 5

So the answer is choice D. This is the only ordered pair that comes from the given table. Something like (2,4) isn't true because the quarterback threw 1 touchdown pass (not 4) in game two.
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A Jerome's rock band recorded 13 songs during an afternoon recording session. Each song took 1/6 of a side of a tape to record.
Lapatulllka [165]
Answer:
number of tape sides = 2.1667 tape sides

Explanation:
We know that each song took 1/6 of a side of tape to be recorded.
To know how many tape sides were used to record 13 songs, all we have to do is cross multiplication as follows:
1 song ..................> 1/6 of a tape side
13 songs ..............> ?? tape sides

Number of tape sides = [(13) * (1/6)] / 1
number of tape sides = 2.1667 tape sides

Hope this helps :)
8 0
3 years ago
Determine whether the statements in (a) and (b) are logically equivalent.
Andru [333]

Answer:

The two statements are logically equivalent.

Step-by-step explanation:

Let X be the statement that Bob is a double math and computer science major.

Y be the statement that Ann is a maths major.

Z be the statement that Ann is a double maths and computer science major.

The two statements written in terms of X, Y and Z now

a. Bob is a double math and computer science major and Ann is a math major, but Ann is not a double math and computer science major.

(x and y) and not z

b. It is not the case that both Bob and Ann are double math and computer science majors, but it is the case that Ann is a math major and Bob is a double math and computer science major.

not (x and z) and (x and y)

Noting that for logical statements,

Negation is represented by ~

And is represented by conjunction sign Λ

Or is represented by disjunction sign V

(x and y) and not z

(x Λ y) Λ (~z)

not (x and z) and (x and y)

~(x Λ z) Λ (x Λ y)

We can then simplify the second statement to obtain the first statement and prove the equivalence of both sides

~(x Λ z) Λ (x Λ y)

Using DE MORGAN'S theory, ~(x Λ z) = (~x) V (~z)

~(x Λ z) Λ (x Λ y) = ((~x) V (~z)) Λ (x Λ y)

Then applying the distributive law to the expression, we can open the bracket up

((~x) V (~z)) Λ (x Λ y)

= ((~x) Λ (x Λ y)) V ((~z) Λ (x Λ y))

Opening the first bracket up further

((~x) Λ (x Λ y)) V ((~z) Λ (x Λ y))

= ((~x) Λ x) Λ y) V ((~z) Λ (x Λ y))

The NEGATION law shows that (~x Λ x) = c (where c is a negation law parameter for when two opposite statements are combined in this manner, it works like a 0 in operation)

((~x) Λ x) Λ y) V ((~z) Λ (x Λ y))

= (c Λ y) V ((~z) Λ (x Λ y))

But (c Λ y) = (y Λ c) = c (according to the UNIVERSALLY BOUND law, see how c works like a 0 now?)

(c Λ y) V ((~z) Λ (x Λ y))

= c V ((~z) Λ (x Λ y))

= ((~z) Λ (x Λ y)) V c (commutative)

And one of the foremost IDENTITY laws is that (any statement) V c = c

((~z) Λ (x Λ y)) V c

= ((~z) Λ (x Λ y))

= (x Λ y) Λ (~z)

Which is the same as the first statement!

PROVED!!!

Hope this Helps!!!

5 0
3 years ago
Read 2 more answers
An arithmetic progression is a sequence of numbers in which the distance (or difference) between any two successive numbers if t
notsponge [240]

Answer:[m, m+d, m+2d, - - - - -, n]

Step-by-step explanation:

We know the formula for arithmetic progression is a_(n) = a_(1) + (n-1)d

Where a_(n) is the nth term of the sequence

a_(1) is the first term of the sequence

n is the number of the term like if we are talking about 7th term so the n is 7.

d is the difference between two successive terms.

For this problem we know our first term that is m, our last term that is n and our difference that is d.

For second term we will use the formula

a_(2) = m + (2-1)d

a_(2) = m + (1)d

a_(2) = m + d

Similarly,

a_(3) = m + (3-1)d

a_(3) = m + (2)d

a_(3) = m + 2d

3 0
3 years ago
Identify the Asymptotes for the graphs below help pls!
scoray [572]
Asymptotes are lines that continues to approach a given curve but never actually meets it! For the second figure it is the Y axis since the curves slowly appear to approach the Y axis! Not sure about the first figure(probably the X axis)!
5 0
2 years ago
Read 2 more answers
Whoever checks my work and tell what I did wrong and the right answer will
gulaghasi [49]

Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolatingd_{1}, we need to get rid of the 1/2 first. Let's multiply each side by 2:

2*m = 2*\frac{1}{2}(d_{1}+  d_{2}) \\2m = d_{1}+  d_{2}

After this, we just subtract d_{2} from each side to get d_{1}=2m- d_{2}. Dark Blue is correct! Let's now plug in those numbers below:

d_{1}=2(10)- 13 = 20-13=7

G: Let's isolate the vw^2 on one side by subtracting y from each side:

vw^{2} +y-y=x-y\\vw^{2}=x-y

Let's now divide each side by v, then put each side under a square root to get our final answer:

\frac{vw^2}{v} = \frac{x-y}{v}\\  w^2 =  \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}

Orange is correct! Again, let's solve the problem underneath:

w=w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

2a-2a+3c=17a-2a+21\\3c = 15a+21\\\frac{3c}{3} = \frac{15a+21}{3}\\c = \frac{15a}{3}+ \frac{21}{3} \\c = 5a+7

Purple is correct! Let's solve the problem:

c = 5(\frac{16}{5})+7 = 16+7 = 23

7 0
2 years ago
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