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3 years ago

Round 123,949 to the underlined point. 4 is highlighted

Mathematics

2 answers:

klemol [59]3 years ago

8 0

Answer:

The number is 123949 (or 123.949 in case comma stands for decimal)

According the rounding rule

if the number to the right is ≥ 5, round up the number to the left, otherwise leave as is, if decimal the number to the right drops, if whole number it is replaced with zero

Round to the number 4, the second number from the right:

123950

123.95

ehidna [41]3 years ago

5 0

$\\ \sf\longmapsto 123,949$

9 is greater than 5

$\\ \sf\longmapsto 123,95$

$\\ \sf\longmapsto 123,9$

9 is greater than 5

$\\ \sf\longmapsto 124$

4 is less than 5

$\\ \sf\longmapsto 12$

2 is less than 5

$\\ \sf\longmapsto 1$

Note

Write the number until underlined point.I did it till end

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There is a 70 percent chance that an airline passenger will check bags. In the next 16 passengers that check in for their flight

GenaCL600 [577]

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a) $P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

b) $P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

c) $P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=17, p=0.7)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

And we want to find this probability:

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

Part b fewer than 10 will check bags

We want this probability:

$P(X$

We can use the complement rule and we have:

$P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

Part c at least 10 bags

We can find this probability like this:

$P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

4 0

3 years ago