360-357+354-351+...+300-297

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

7 0

$360-357+354-351+...+300-297\\\\=(360-357)+(354-351)+...+(300-297)\\\\=\underbrace{3+3+...+3}_{11}=11\times3=33$

Lapatulllka [165]3 years ago

5 0

$x=360-357+354-351+...+300-297=3+3+3+...+3=3n\\\\\ \ \ \ and \ \ \ 300=360-6\cdot(n-1)\ \ \ \Rightarrow\ \ \ 6n=66\ /:6\ \ \ \Rightarrow\ \ \ n=11\\\\x=3\cdot11=33$

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ivolga24 [154]

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3 years ago

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baherus [9]

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In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.553, n = 10$

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This is

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$