Answer:
Jamal's answer isn't reasonable because the sum of 183 and 198 is 381, which is way more than 300 and nowhere less than 300.
Step-by-step explanation:
Jamal makes an assertion that the sum of 183 and 198 is less than 300.
We are to check if Jamal's answer is reasonable or not.
183 + 198 = 381 > 300
The sum of the two numbers, 381, is evidently not less than 300, hence, Jamal's answer isn't reasonable because it is downright wrong.
Hope this Helps!!!
Answer:
a
Step-by-step explanation:
good luck :)
1, –21.8, –56
–5.7, –21.8, –51.3
1, –16.1, –50.3
0, –17.1, –51.3
If someone answers this, i'd like if someone could help me to know how to solve it too, please and thank you!
1, - 16.1, - 50.3 ← third on list
To find the required terms, substitute n = 1, 4, 10 into the rule and evaluate
A(1) = 1 + (1 - 1)(- 5.7) = 1 + 0 = 1
A(4) = 1 + (4 - 1)(- 5.7 ) = 1 + (3 × - 5.7 ) = 1 - 17.1 = - 16.1
A(10) = 1 + (10 - 1 )(- 5.7) = 1 + (9 × - 5.7) = 1 - 51.3 = - 50.3
Answer:
8 orders in 1 day
Step-by-step explanation:
80 divided by 8 is 10. 8 divided by 8 is 1.
Answer:
See explanation below
Step-by-step explanation:
Here a coin was tossed three times.
Let H = head & T = tail
Find the following:
a) The sample space:
Since a coin is tossed thrice, all possible outcome would be:
S = { HHH, HHT, HTH, HTT, TTT, TTH, THH, THT}
b) i) A = Exactly 2 tails: Here exactly 2 tails were recorded.
A = {HTT, TTH, THT}
ii) B = at least two tails: Here 2 or more tails were recorded.
B = {HTT, TTT, TTH, THT}
iii) C = the last two tosses are heads:
C = { HHH, THH}
c) List the elements of the following events:
i) A. This means all outcomes in A
= {HTT, TTH, THT}
ii) A∪B. A union B, means all possible outcomes present in A or B or in both
= {HTT, TTH, THT, TTT}
iii) A∩B. This means all possible outcomes of A that are present in B.
= {HTT, TTH, THT}
iv) A∩C. All outcomes A that are present in B
= {∅}
The sample space of tossing a coin three times consists of eight possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Events A, B, and C can be determined by listing the appropriate outcomes. The intersection and union of events A and B can also be determined.
(a) The sample space, Ω, of tossing a coin three times can be determined by listing all the possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.
(b) i. A = {HHT, HTH, THH}
ii. B = {TTT, TTH, THT, HTT, HHT, HTH, THH}
iii. C = {HTH, TTH}
(c) i. A = {HHT, HTH, THH}
ii. A∪B = {HHT, HTH, THH, TTT, TTH, THT, HTT, HHT}
iii. A∩B = {HHT, HTH, THH}
iv. A∩C = {HHT, HTH}
#SPJ3
What is the solution to the story?
13.80
10.60
07.00
8.25
Answer:
C. $7.00
Step-by-step explanation:
8 times 7 is 56 + 5 = 61
Answer:
0
Step-by-step explanation:
0 is the probability of a die coming up with the number 10
because a die contains numbers from 1 to 6 so probability of numbers coming between are possible but 10 is not possible