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viktelen [127]
3 years ago
10

Write a rule for each sequence. Find the next three terms. 8,14,20,26

Mathematics
1 answer:
anastassius [24]3 years ago
8 0
Aruthmetic sequene is
an=a1+(n-1)d
where d=common difference between terms
adds 6 every time
d=6
first term is 8
a1=8
8+6(n-1)
distribute
8+6n-6
8-6+6n
2+6n is answer
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Use a calculator to find the value of the acute 0 to the nearest degree<br> Tan0=7.7336
Mademuasel [1]

In degree mode, the answer is around 83. You use the inverse tangent button on the calculator to get rid of the tangent and find what theta is (make sure your calculator is in degree mode)

6 0
3 years ago
An arithmetic sequence is represented by the explicit formula A(n) = 2 + 9(n - 1). What is the recursive formula?. . A. A(n) = A
KiRa [710]
A(n) = 2 + 9(n - 1) = 2 + 9n - 9
A(n + 1) = 2 + 9(n + 1 - 1) = 2 + 9n = (2 + 9n - 9) + 9 = A(n) + 9

Therefore, the recursive formular is A(n) = A(n - 1) + 9
6 0
3 years ago
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Jane wants to estimate the proportion of students on her campus who eat cauliflower. after surveying 39 ​students, she finds 4 w
stellarik [79]

Let x be the number of students who eat cauliflower.

Therefore, x=4.

Let n be the total number of students surveyed.

Therefore, n=39

Thus, \hat p=\frac{4}{39} =0.10256

Now, for 90% confidence level, from the table we know that Z=1.645.

The formula for the interval range of proportion of students is :

p= \hat p\pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}}

Plugging in the values we get:

p=0.10256\pm 1.645\sqrt{\frac{0.10256(1-0.10256)}{39}}=0.10256\pm 0.04858=0.15114, 0.05398

Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).

7 0
4 years ago
Plz Help
kondor19780726 [428]

Answer:anna

Step-by-step explanation: becase well he got the most $ added in the 2cd day

3 0
3 years ago
How to find r in this equation using combination formula C(8,r)=28<br>​
slavikrds [6]

Answer:

r = 2

Step-by-step explanation:

We have the formula of ^nC_r = \frac{n!}{r! (n-r)!}

Now, it is given that ^8C_r = \frac{8!}{r! (8-r)!} = 28 ........ (1)

And we have to find the value of r which satisfy the above equation.

So, r! (8-r)! = \frac{8!}{28} = \frac{40320}{28} = 1440

Now, we have to use the trial method to find the value of r.

For r = 1, 1! (8-1)! = 7! = 5040 \neq 1440

Hence, r can not be 1.

Now, put r = 2, 2! (8-2)! = 2 \times 6! = 1440

Therefore, r = 2 (Answer)

5 0
3 years ago
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