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just olya [345]
3 years ago
5

Average annual rainfall totals for cities in new York

Mathematics
1 answer:
Archy [21]3 years ago
8 0
The average annual rainfall totals for cities in new york is 43 inches

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What is one way to test whether an unknown solution is acidic or basic?
jasenka [17]

Answer:

Bases change red litmus into blue and bases have a slippery and soapy texture. Therefore, we can conclude that out of the given options, one way to test whether an unknown solution is acidic or basic is to check whether the solution has a slippery feel

Step-by-step explanation:

3 0
3 years ago
By selling a watch in Rs 150 and Rs 200 loss and profit are happened
Travka [436]

Answer:

Rs 175

Step-by-step explanation:

Suppose the cost is x and at Rs150 the loss is 150-x (this should be a negative number).

At Rs200, the profit is 200-x.

So we have an equation: minus 150 minus x is equal to 200 minus x.

To solve the equation, the cost price X is Rs175.

3 0
3 years ago
The graph below shows the amount of money Janet earned at her new job.
navik [9.2K]

Answer:

Janets earns $7.50 an hour.

Step-by-step explanation:

2 hours : 15 dollars

divide by 2

1 hour : 7.50 dollars

Hope dis helped!

6 0
3 years ago
Read 2 more answers
In a group of 20 students, half have blue eyes, 10% have brown eyes, and the rest have hazel eyes. How many students have hazel
Ber [7]
So you have 20 people and 10 percent have brown eyes. So if 100% is 20, to find 10 percent you divide it by 10 to get 10% is 2. You know all the rest have hazel eyes so you just do 20-2 to get 18. Therefore 18 people have hazel eyes. Hope this helps :)
5 0
3 years ago
Consider a Poisson distribution with μ = 6.
bearhunter [10]

Answer:

a) P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b) f(2) = 0.04462

c) f(1) = 0.01487

d) P(X \geq 2) = 0.93803

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of successes

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

In this question:

\mu = 6

a. Write the appropriate Poisson probability function.

Considering \mu = 6

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b. Compute f (2).

This is P(X = 2). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

Then

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803

So

P(X \geq 2) = 0.93803

5 0
3 years ago
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