Answer:
The inference that can be made using the dot plot is:
The range of round 1 is greater than the round 2 range.
Step-by-step explanation:
<u>Round 1:</u>
Score Frequency
1 0
2 2
3 3
4 2
5 1
Hence, the minimum score of Round 1 is: 2
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-2
=3
Similarly, <u>Round-2</u>
Score Frequency
1 0
2 0
3 0
4 4
5 4
Hence, the minimum score of Round 1 is: 4
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-4
=1
The scores of round 2 are higher than round-1.
Since round 2 have a higher frequency for higher scores as compared to round-1.
Hence, Range of round 1 is greater than the range of Round-2.
Answer:
Mia scored 6 baskets Last Saturday.
Step-by-step explanation:
Let:
Mia made score of baskets on Last Saturday = x
Mia made score of baskets yesterday = 3x (Mia made three times as many baskets in yesterday’s basketball game as she did last Saturday. three times means multiply 3 with score of Last Saturday baskets, i.e. x)
Total score of baskets = 24
We need to find How many baskets did Mia score last Saturday?
We can write:
<em>Total Score = Score on Last Saturday + Score yester</em>day
Solving this equation we can find value of x:
So, we get value of x = 6
Mia made score of baskets on Last Saturday = x = 6
Mia scored 6 baskets Last Saturday.
1/2 + 9/20 = 19/20
1/20 + 1/10 = 3/20
1/8 + 11/16 = 13/16
2/9 + 5/9 = 7/9
3/20 + 3/4 = 18/20 = 9/10
7/17 + 1/17 = 8/17
Part A
Answer: The common ratio is -2
-----------------------------------
Explanation:
To get the common ratio r, we divide any term by the previous one
One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2
Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time
Side Note: each term is multiplied by -2 to get the next term
============================================================
Part B
Answer:
The rule for the sequence is
a(n) = (-2)^(n-1)
where n starts at n = 1
-----------------------------------
Explanation:
Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio
The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)
============================================================
Part C
Answer: The next three terms are 16, -32, 64
-----------------------------------
Explanation:
We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms
-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64
showing that the next three terms are 16, -32, and 64
An alternative is to use the formula found in part B
Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 .... which matches with what we got earlier
Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 .... which matches with what we got earlier
Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 .... which matches with what we got earlier
while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)
Answer:
because it may not be suitable for all the time
