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Mice21 [21]
4 years ago
13

What is the sum if the first 22 terms of the following series? 86 + 79 + 72 + 65 +58+...

Mathematics
1 answer:
Arturiano [62]4 years ago
5 0
Seems to be an arythmetic sequence
Sn=[n(a1+an)]/2
where
Sn means sum of all terms up the nth term
n=number of terms
a1=first term
an=nth term

so from 86 to the 22th term is from a1 to a22

find teh sequence
miknus 7 each time
an=a1+d(n-1)
an=87-7(n-1)
find 22n term
a22=87-7(22-1)
a22=87-7(21)
a22=87-147
a22=-60

S22=[22(87-60)]/2
S22=[22(27)]/2
S22=594/2
S22=297

the sum is 297

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a random sample of 4 claims are selected from a lot of 12 that has 3 nonconforming units. using the hypergeometric distribution
Sloan [31]

Answer:

The probability that the sample will contain exactly 0 nonconforming units is P=0.25.

The probability that the sample will contain exactly 1 nonconforming units is P=0.51.

.

Step-by-step explanation:

We have a sample of size n=4, taken out of a lot of N=12 units, where K=3 are non-conforming units.

We can write the probability mass function as:

P(x=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

where k is the number of non-conforming units on the sample of n=4.

We can calculate the probability of getting no non-conforming units (k=0) as:

P(x=0)=\frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}}=\frac{1*126}{495}=\frac{126}{495} = 0.25

We can calculate the probability of getting one non-conforming units (k=1) as:

P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}=\frac{3*84}{495}=\frac{252}{495} = 0.51

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4 years ago
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