All categories

3 years ago

Determine the relationship described in each table

Mathematics

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

Written below

Step-by-step explanation:

The first table is linear

the Y differences are +4 going down the table

Second is exponential

the Y differences are divided by 2 going down the table

The third is quadratic

The Y differences are first -14, -6, and 2 but then you take those differences to get the differences of +8 down the table

You might be interested in

How do you do this problem

Ainat [17]

You divide (1354.5 kg of potatoes) by (12 stores in the Bronx).

The quotient is ________ kg of potatoes per store in the Bronx.

The division is easy to do with your pencil or your calculator.

7 0

3 years ago

Shania is buying a car. She has narrowed her choice down to three cars: a sedan, a hatchback, and a wagon. She uses a spinner di

Bess [88]

Answer:

its D

Step-by-step explanation:

8 0

3 years ago

Read 3 more answers

Solve using the pathagorean theorem

Sergio039 [100]

Answer:

A^(2) + B^(2) = C^(2)?????????????

Step-by-step explanation:

that's an isosceles triangle, not a right-angle triangle

8 0

2 years ago

As Español

goldenfox [79]

Answer:

Step-by-step explanation:

120-73 = S

therefore 73+s = 120

so 120- s = 73

5 0

2 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

Read 2 more answers