which sequences are arithmetic? check all that apply. –8.6, –5.0, –1.4, 2.2, 5.8, … 2, –2.2, 2.42, –2.662, 2.9282, … 5, 1, –3, –
sergiy2304 [10]
In an arithmetic sequence, the next term is found by adding a constant term to each number to arrive at the next number. The common difference can be found by subtracting the first term from the second term.
-8.6, -5.0, -1.4, 2.2, 5.8....the common difference here is 3.6
-8.6 + 3.6 = -5.0
-5.0 + 3.6 = -1.4
1.4 + 3.6 = 2.2
2.2 + 3.6 = 5.8
so this IS an arithmetic sequence.
2,-2.2, 2.42, -2.662, 2.9282...there is no common difference..so this is not an arithmetic sequence
5,1,-3,-7,-11....common difference is -4
5 + (-4) = 1
1 + (-4) = -3
-3 + (-4) = -7
-7 + (-4) = -11
this IS an arithmetic sequence
-3,3,9,15,21...common difference is 6
-3 + 6 = 3
3 + 6 = 9
9 + 6 = 15
15 + 6 = 21
this IS an arithmetic sequence
-6.2, -3.1, -1.55, -0.775, -0.3875...this is not an arithmetic sequence
The answer is A I’m pretty sure
Answer:
Here's What you do:
- Draw a line that comes straight through the middle
- It has to create 4 - 90° angles (use a protractor to measure)
There's a picture attached on how a perpendicular lines should look like.
Step-by-step explanation:
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-Chetan K
Answer:
The probability that the intersection will come under the emergency program is 0.1587.
Step-by-step explanation:
Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.
Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.
For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.
Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).

The values of
, the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program

Answer:
1
Step-by-step explanation:
-20 divided by -2 is the same as 20 divided by 2, which is 10. 11-10=1 so the answer is 1.