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1 year ago

12

The number of hours walked varies inversely with the speed of the walker. if it takes sam 12 hours to complete his walking goal

at 5 miles per hour, how long would it take him at 3 miles per hour?

1 answer:

Effectus [21]1 year ago

7 0

We let k be the proportionality constant for the relationship between number of hours, h and speed of the walker, s.

h = k/s
Substituting the known values,
12 = k/5
k = 60

For the second scenario,
h = k/s
Substituting the calculated value for k and the given value for speed,
h = (60)(3 miles/hour)
h = 20 hours
h = 20 hours

Therefore, it will take 20 hours to walk with a speed of 3 miles per hour.

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The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.8. Answer the followi

slava [35]

Answer:

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

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

b) $P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

Step-by-step explanation:

Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that $X \sim Poisson(\lambda=0.8)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda=0.8$

$E(X)=\mu =\lambda=0.8$

Part a

For this case we want this probability: $P(X>2)$

And for this case we can use the complement rule like this:

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

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

Part b

Using the probability mass function we have:

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

3 0

2 years ago

A spinner with 4 congruent sectors labeled 1-4 is spun. Then a die is rolled. What is the probability of getting even numbers on

SOVA2 [1]

The spinner is divided into 4 equal sections number 1 to 4.

So, for spinner, total sections = 4
Favorable sections = 2 (i.e sections with even numbers)

So, probability of getting even number on the spinner = 2/4 = 1/2

Total number of outcome when a dice is rolled = 6
Favorable outcomes= 3 (i.e outcomes with 2,4 and 6)

So, probability of getting an even number = 3/6 = 1/2

Since both events are independent, we can write:

The probability of getting an even number in both events = 1/2 x 1/2 = 1/4

4 0

2 years ago

A stadium asked for $760,000 for naming rights. During a bidding war for the stadium naming rights, one company bid 7% more than

alex41 [277]

Answer:

$\$\ 829464$

Step-by-step explanation:

Bid of first company:

Bid is $7\%$ more

$7\%$ of $760000=\frac{7}{100}\times760000=53200$

Bid of first $=760000+53200=813200$

Bid of second company:

Bid of second is $2\%$ greater than that of first

$2\%$ of first bid $=\frac{2}{100}\times 813200=16264$

Hence Bid of second company $=813200+16264=829464$

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2 years ago

Which expression is equivalent to i 233

Fiesta28 [93]

Start with second, third and fourth degree of imaginary unit i:

$i^2=-1, \\ i^3=i^2\cdot i=-1\cdot i=-i, \\ i^4=i^2\cdot i^2=-1\cdot (-1)=1$ .

Since 233=232+1=4·58+1, then $i^{233}=i^{4\cdot 58+1}=(i^4)^{58}\cdot i^1=1^{58}\cdot i=i$ .

Answer: $i^{233}=i$

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2 years ago

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Prove that x is a subset of y then x union z is a subset of y union z for all sets x y and z

Ganezh [65]

We are to show that if X ⊆ Y then (X ∪ Z) ⊆ (Y ∪ Z) for sets X, Y, Z.

Assume that a is a representative element of X, that is, a ∈ X. By the definition of union, a ∈ X ∪ Z. Now because X ⊆ Y and we assumed a ∈ X, then a ∈ Y by the definition of subset. And because a ∈ Y, then a ∈ Y ∪ Z by definition of union.

We chose our representative element, a, and showed that a ∈ X ∪ Y implies that a ∈ Y ∪ Z and this completes the proof.

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1 year ago