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Aleks04 [339]
3 years ago
12

Find the 15th term of the arithmetic sequence 13,8,3,-2

Mathematics
1 answer:
Lunna [17]3 years ago
4 0

Answer:

-57

Step-by-step explanation:

13,8,3,-2 (decreasing by 5)

13,8,3,-2,-7,-12,-17,-22,-27,-32,-37,-42,-47,-52,-57.

The 15th term is -57.

<u><em>Ace</em></u>

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How do you find the radius of a right cylinder with having the height and surface area?
RoseWind [281]
So we know that surface area is top circle +bottom circle+ (circumfurence times height so) 2(π(radius)^2)+(height(2π(r))) so fill in the blanks
it's kind of hard to see but πr means pi times radius (r)
2πr^2+((4.2)(2)(π)(r))=628.3
factor out the 2 so  the equation would mean
2((πr^2)+(4.2πr))=628.3
divide both sides by 2
πr^2+4.2πr=314.15
what's interesting is that the first few digits of pi are 3.1415 so we aproximate pi as 3.14 and divide both sides by 3.1415
r^2+4.2r=100
subtract 100 from both sides
r^2+4.2r-100=0
factor
(x-8.11812)(x+12.3181)=0
so (if xy=0 then x or/and y must be 0 so)we set them to 0 so
x-8.11812=0
x=8.11812

second one
x+12.3181=0
x=-12.3181
since radii (or any legnth) cannot be negative, we can cross this solution out so

radius=8.11812
5 0
4 years ago
9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions
just olya [345]

Answer:

\frac{7}{30}\approx0.23

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5

Now, probability of getting the second question as easy question is given as:

P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}

Now, probability that the first two contestants will get easy questions is given by the product of P(E1)\ and\ P(E2). So,

P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23

Therefore, the probability that the first two contestants will get easy questions is \frac{7}{30}\approx0.23

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3 years ago
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Answer:

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