If an arithmetic series has a1 equals=44, d equals 5, and n=24, what is sn?
1 answer:
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Answer:
I usually explain even in the answer.
So we use something called distance formula which is branched of Pythagorean theorem.
But we dont need to as it just makes it more complicated. We need to find split into 2 vectors, one vertical, and horizontal. The horizontal is 5 long.
The vertical is 1, you can find them by calculating how long is it and how tall.
Use pythogorean theorem from the formula and do 5^2+1^2 = c^2
25+1 = 26, so the answer is √26
I am pretty sure answer is right. Always take abolute value
<u>√26</u>
Answer:
30500 = 3.05·10^4
Step-by-step explanation:
Your calculator can do this for you. You may need to set the display to scientific notation, if that's the form of the answer you want.
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This can be computed by converting both numbers to standard form:
(5·10^2) +(3·10^4)
= 500 +30000 = 30500 = 3.05·10^4
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Addition of numbers in scientific notation in general requires that they have the same power of 10. It may be convenient to convert both numbers to the highest power of 10.
5·10^2 + 3·10^4
= 0.05·10^4 +3·10^4 . . . . now both have multipliers of 10^4
= (0.05 +3)·10^4
= 3.05·10^4
Answer:
The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.
Step-by-step explanation:
Denote the different kinds of drivers as follows:
L = low-risk drivers
M = moderate-risk drivers
H = high-risk drivers
The information provided is:
P (L) = 0.50
P (M) = 0.30
P (H) = 0.20
Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.
The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:
S = {HHHH, HHHL, HHHM, HHMM}
Compute the probability of the combination {HHHH} as follows:
P (HHHH) = [P (H)]⁴
= [0.20]⁴
= 0.0016
Compute the probability of the combination {HHHL} as follows:
P (HHHL) = × [P (H)]³ × P (L)
= 4 × (0.20)³ × 0.50
= 0.016
Compute the probability of the combination {HHHM} as follows:
P (HHHL) = × [P (H)]³ × P (M)
= 4 × (0.20)³ × 0.30
= 0.0096
Compute the probability of the combination {HHMM} as follows:
P (HHMM) = × [P (H)]² × [P (M)]²
= 6 × (0.20)² × (0.30)²
= 0.0216
Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:
P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)
+ P (HHMM)
= 0.0016 + 0.016 + 0.0096 + 0.0216
= 0.0488
Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.