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3 years ago

Members of a softball team raised $1319.25 to go to a tournament. They rented a bus

Mathematics

1 answer:

harina [27]3 years ago

7 0

Answer:

13 players

Step-by-step explanation:

Money Raised=RentedBus+Meals

1319.25=776.50+41.75p (p is for the number of players)(Now we solve for x)

Subtract 776.50 from both sides: 542.75=41.75x

Divide both sides by 41.75: 13=x

Remember that x is for the number of players so there'll be 13 players going to the tournament.

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A die is tossed. If the number rolled is between 2 and 5, inclusive, what is the probability the number rolled is 4?

lesya [120]

<h2>The possibility is Likely.</h2>

5 0

3 years ago

Tanisha's mother buys 38 lemons. They know it takes 8 lemons to make 80 oz. of lemonade. Does Tanisha need more lemons? If not,

Cloud [144]

She has extra because she only wants to make 80oz and it takes 8 and she has 30 so 30 is extra

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3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

4 years ago

3. company pays a greater annual salary? Show your Company A offers a semimonthly salary of $2432, and Company B offers a biweek

Eduardwww [97]

Company A offer
2,432×24(semimonthly)
=58,368 per year

Company B offer
2,390×26 (biweekly)
=62,140 per year

So company B pays a greater annual salary

Hope it helps!

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4 years ago

Samuel can type nearly 40 words per minute. Use this information to find the number of hours it would take him to type 3.8 × 105

lora16 [44]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers