Consider this system of equations. Which equation represents the first equation written in slope-intercept form?
2 answers:
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<h2>Answer:</h2>
1. A coin is tossed four times. What is the probability of getting four heads?
Each toss has a 1/2 chance of getting a head.
So the chance of getting all four heads can be calculated as :
2. A coin is tossed four times. What is the probability of getting two heads?
Each toss can have 2 results, so 4 flips will have or 16 results.
Getting two heads means getting two tails also. So, we can get the number of times two heads come = = 6
We can write the groups like - {HHTT,HTHT,HTTH,THHT,THTH,TTHH}
So, the required probability is : 6/16 or 3/8.
3. Not getting two heads means getting 3 tails and 1 head or all tails.
Probability of having all tails =
Probability of one head(1st trial) and three tails = 1/16
Probability of one head (2nd trial) and three tails (the 1st, 3rd and 4th trials) = 1/16
Probability of one head (3rd trial) and three tails (the 1st, 2nd and 4th trials) = 1/16
Probability of one head (4th trial) and three tails (the 1st, 2nd and 3rd trials) = 1/16
So, the total probability showing only one or none head and at least three tails =
Answer:
Step-by-step explanation:
NA = √[(- 4 - 1 )² + (- 3 - 2)²] = 5√2
AT = √[(8 - 1 )² + (1 - 2)²] = 5√2
TS = √[(3 - 8 )² + (- 4 - 1)²] = 5√2
NS = √[(- 4 - 3 )² + (- 3 + 4)²] = 5√2
NA = AT = TS = NS = 5√2
= (- 3 - 2) / (- 4 - 1) = 1 ........ <em>(1)</em>
= (- 4 - 1) / (3 - 8 ) = 1 ......... <em>(2)</em>
From (1) and (2) ⇒ NA║TS
= ( 1 - 2) / ( 8 - 1) = - 1 / 7 .......... <em>(3)</em>
= ( - 4 + 3) / ( 3 + 4) = - 1 / 7 .... <em>(4)</em>
From (3) and (4) ⇒ AT║NS
Thus, NATS is rhombus.