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Hatshy [7]
2 years ago
14

Match the vocab terms with the correct definition.

Mathematics
2 answers:
Anna007 [38]2 years ago
8 0
7. compound events that that are not affected by other's outcomes
pishuonlain [190]2 years ago
7 0
1 is theoretical probability
2 is overlapping events
3 is combination
4 is dependent events
5 is disjointed events
6 is permutation
7 is independent events
8 is experimental probability
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Subtract 6 from me. Then multiply by 2. If you subtract 40 and then divide by 4, you get 8. What number am I?
aliina [53]

Answer:

42

Step-by-step explanation:

Let the number be "x"

We translate the word equation to algebraic equation.

First,

subtract 6 from me, so we have:

x - 6

Now,

Multiply by 2, so we have:

2(x-6)

Now,

Subtract 40, then divide by 4:

\frac{2(x-6)-40}{4}

Now, you get "8", so equate this to 8, we get:

\frac{2(x-6)-40}{4}=8

We now solve using algebra:

\frac{2(x-6)-40}{4}=8\\2(x-6)-40=4*8\\2(x-6)-40=32\\2x-12-40=32\\2x-52=32\\2x=32+52\\2x=84\\x=42

The number is 42

7 0
3 years ago
7th term in geometric sequence an=2 5(n-1)
Soloha48 [4]
To find the 7th term, all you have to do is plug it in the equation.

First n would equal 7 because we are looking for the 7th term.

Now, let's plug everything we know into the equation.

a7 = 2+5 * (7-1)

= 2 + 5 * 6

= 2 + 30

= 32

In conclusion, the 7th term would equal 32.
3 0
3 years ago
A fair die is cast four times. Calculate
svetlana [45]

Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
8 0
3 years ago
Based on the pattern of the drawings, which conjecture
anzhelika [568]

Answer:

The most reasonable answer would likely be the first.

"When a pair of parallel lines is intersected by a third line, the alternate exterior angles are congruent."

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help someone if you not sleep please
mart [117]

Hello from MrBillDoesMath!

Answer:

+\- 10i

Discussion:

Fortunately, I can do this one in my sleep...... );


+\- sqrt(-100)  =

+\- 10 i                         where i = sqrt(-1)


Thank you,

MrB

8 0
2 years ago
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