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2 years ago

Two trains leave stations

Mathematics

2 answers:

gavmur [86]2 years ago

7 0

Roughly one and a half hours in

Nat2105 [25]2 years ago

4 0

About 1 and a half hours

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Which graph represents the equation y = 3/4 x - 3

Kazeer [188]

Answer:

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2 years ago

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brilliants [131]

Answer:

Step-by-step explanation:

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2 years ago

25% of all who enters a race do not complete. 30 haveentered.

nikklg [1K]

Answer:

The probability that exactly 5 are unable to complete the race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the race?

So, We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

p is the probability of success i.e. 25% = 0.25

q is the probability of failure = 1- p = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the race

n = 30

r = 5

$P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}$

$P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}$

$P(X=5) =0.1047$

Hence the probability that exactly 5 are unable to complete the race is 0.1047

6 0

3 years ago

If DE≅EF, CD = v + 45, and CF = 4v, what is the value of v?

irinina [24]

Answer:

<h2>since DE≅EF</h2>

CD=CF (cpct)

$v + 45 = 4v \\ 4v - v = 45 \\ 3v = 45 \\ v = \frac{45}{3} \\ v = 15$

<h2>hope it helps you<3</h2>

3 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

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

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

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

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".

Solution to the problem

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

$X \sim Binom(n=20, p=0.2)$

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

We want this probability:

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

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

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

2 years ago