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3 years ago

The next number in the arithmetic sequence 10,23,36,? Is

Mathematics

1 answer:

rjkz [21]3 years ago

6 0

It would be 49
10+13=23
23+13=36
36+13=49
I think this is right

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What is 3×584 in expanded form

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Three mulitplied by five-hundred eighty four

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You are buying tickets to a school event. You may purchase at most 8 tickets to receive a discount. Write an inequality to expre

podryga [215]

8 is greater than or equal to x
(where x = the number of tickets)

or x is less than or equal to 8

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3 years ago

the left and right page numbers of an open book are two consecutive integers whose sum is 589. find these page numbers

kramer

Answer:

267 and 322

Step-by-step explanation:

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Solve each equation for the indicated variable. u = -3a - 3 , for a

MAXImum [283]

Answer:

$a = \frac{u + 3}{-3}$

Step-by-step explanation:

Isolate the variable, a. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

PEMDAS is the order of operation and stands for:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

First, add 3 to both sides of the equation:

u = -3a - 3

u (+3) = -3a - 3 (+3)

u + 3 = -3a

Next, divide -3 from both sides of the equation:

(u + 3)/-3 = (-3a)/-3

a = (u + 3)/-3

$a = \frac{u + 3}{-3}$ is your answer.

8 0

3 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

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

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

$\ln{e^{-0.01x}} = \ln{0.08}$

$-0.01x = \ln{0.08}$

$x = -\frac{\ln{0.08}}{0.01}$

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago