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4 years ago

A hellium filled blimp stays aloft

Mathematics

2 answers:

Ahat [919]4 years ago

8 0

Answer:

thats not a qustion im sorry i cannot answer it

Step-by-step explanation:

SpyIntel [72]4 years ago

3 0

I mean pretty much.

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The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupie

nikitadnepr [17]

Answer:

a) $P(X=3)$

And we can use the probability mass function and we got:

$P(X=3) = (10C3) (0.4)^3 (1-0.4)^{10-3} = 0.215$

b) For this case we want this probability:

$P(X \geq 1)$

and the probability that the one call would be not occupied is 1-0.4=0.6

And if we use the complement rule we got:

$P(X \geq 1)= 1-P(X$

$P(X=0) = (10C0) (0.6)^0 (1-0.6)^{10-0} = 0.000105$

$P(X \geq 1)= 1-P(X$

c) $E(X) = np = 10*0.4 = 4$

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".

a. What is the probability that for exactly three calls, the lines are occupied?

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

$X \sim Binom(n=10, p=0.4)$

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:

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

And for this case we want this probability:

$P(X=3)$

And we can use the probability mass function and we got:

$P(X=3) = (10C3) (0.4)^3 (1-0.4)^{10-3} = 0.215$

b. What is the probability that for at least one call, the lines are not occupied?

For this case we want this probability:

$P(X \geq 1)$

and the probability that the one call would be not occupied is 1-0.4=0.6

And if we use the complement rule we got:

$P(X \geq 1)= 1-P(X$

$P(X=0) = (10C0) (0.6)^0 (1-0.6)^{10-0} = 0.000105$

$P(X \geq 1)= 1-P(X$

c. What is the expected number of calls in which the lines are all occupied?

For this case the expected value is given by:

$E(X) = np = 10*0.4 = 4$

8 0

3 years ago

Juanita Domingo's parents want to establish a college trust for her. They want to make 16 quarterly withdrawals of $2000, with t

kotegsom [21]

Answer:

The amount to be deposited now to provide for this trust is $119,392.16.

Step-by-step explanation:

This problem is based on ordinary annuity.

An ordinary annuity is a sequence of fixed payments made, every consecutive period, over a fixed interval.

The formula to compute ordinary annuity is:

$OA=P[\frac{q^{n}-1}{q^{n}(q-1)}]$

Here <em>qⁿ </em>is:

$q^{n}=1+\frac{r}{Number\ of\ periods}=1+\frac{0.067}{4}=1.01675$

Compute the ordinary annuity as follows:

$OA=P[\frac{q^{n}-1}{q^{n}(q-1)}]=2000\times\frac{(1.01675)^{16}-1}{(1.01675)^{16}[1.01675-1]}=2000\times\frac{0.30445}{0.0051}=119392.16$

Thus, the amount to be deposited now to provide for this trust is $119,392.16.

4 0

3 years ago

If ∠A and ∠B are supplementary angles and m∠A = 3m∠B, find m∠A and m∠B.

rjkz [21]

∠A and ∠B are supplementary so that means they have a sum of 180°.
m ∠A=3(m ∠B) and m ∠B=?

m ∠A + m ∠B=180°
3 (m ∠B) + m ∠B=180°
4 m ∠B=180°
4/4 m ∠B=180°/4
m ∠B=45° and m ∠A=3(m ∠B)=3 (45)=135°

6 0

3 years ago

Someone help 6th grade math i will give brainliest

KengaRu [80]

Answer:

A. 243

B. 1

C. 64

D. 0.36

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need help pls I’m taking a test and I don’t know what to do.

Goryan [66]

B I think this needs to be longer answer but b

4 0

3 years ago