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3 years ago

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A boat traveled 240 miles downstream, then 240 miles back up stream. The trip downstream took 20 hours. The trip back up stream

took 60 hours.
The speed of the boat in still water is ______ miles per hour.

The speed of the current is ______ miles per hour.

2 answers:

MaRussiya [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Rama09 [41]3 years ago

4 0

<u>Answer:</u>

The speed of the boat in still water is <em>8 miles per hour</em>.

The speed of the current is <em>4 miles per hour</em>.

<u>Solution:</u>

We know the distance formula,

$Distance=\frac{speed}{time}$

$\Rightarrow Speed=Distance\times Time$

As boat travelled 240 miles downstream in 20 hours,

speed= $\frac{240}{20} =12$ miles per hour.

As boat travelled 240 miles upstream in 60 hours,

speed= $\frac{240}{60} = 4$ miles per hour.

Let the speed of boat in still water be x and the speed of current be y.

So, the equations formed are: $x+y=12$ (downstream) --- (a) and $x-y=4$ (upstream). --- (b)

On solving, (a)

$\Rightarrow x=12-y$ --- (c)

Substituting (c) in (b), we get $12-y-y=4$

$\Rightarrow 12-2y=4 \Rightarrow 12-4=2y \Rightarrow 8=2y \Rightarrow \frac{8}{2}=y$

Therefore, y=4 --- (d)

On substituting (d) in (a) we get,

$x+4=12 \Rightarrow x=12-4 \Rightarrow x=8$

Therefore, x=8

Hence, Speed of boat in still water= 8 miles per hour and speed of current is 4 miles per hour.

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3 years ago

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

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

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

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

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4 years ago

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seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

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We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago