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3 years ago

Pls solve~~~~~~~~~~~~~~~~~~~~~~~~

Mathematics

1 answer:

IRISSAK [1]3 years ago

4 0

Answer:

Step-by-step explanation:

P°×q3

=1×q3

=q3

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Aurora Corporation operated without insurance coverage for the first month of 2015. Then, on February 1, 2015, the company paid

vitfil [10]

Answer:

$2200

Step-by-step explanation:

$4800 premium for 2 years means for 2* 12 = 24 months, thus each month, the insurance expense is:

4800/24 = 200 dollars

Since they haven't used insurance in January, they will use insurance expense for the rest of the 12 - 1 = 11 months, thus the expense would be:

200 * 100 = $2200

4 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

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 sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago

What is the solution to the equation below. square root x - 4 = -2

ArbitrLikvidat [17]

Answer:

The answer is b.

Step-by-step explanation:

x-4=-2

+4. +4

x= 2

8 0

2 years ago

Read 2 more answers

One tank is filling at a rate of 5/8 gallon per 7/10 hour. A second tank is filling at a rate of 5/9 gallon per 2/3 hour. Which

anzhelika [568]

First recognize that the unit rate we're finding is gallons per hour, or $\frac{g}{h}$ .

rate for Tank 1 = x gallons per hour = a gallons / b hours

$T_{1}=\frac{\frac{5}{8}}{\frac{7}{10}}\\\\T_{1}=\frac{5}{8}*\frac{10}{7}\\\\T_{1}=\frac{50}{56}\\\\\\Fill\ rate\ of\ Tank\ 1=\frac{25}{28}$

rate for Tank 2 = y gallons per hour = c gallons / d hours

$T_{2}=\frac{\frac{5}{9}}{\frac{2}{3}}\\\\T_{2}=\frac{5}{9}*\frac{3}{2}\\\\T_{2}=\frac{15}{18}\\\\T_{2}=\frac{15}{18}\\\\\\Fill\ rate\ of\ Tank\ 2=\frac{5}{6}$