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4 years ago

If 6 cats can kill 6 rats in 6 minutes, how many cats will be needed to kill 100 rats in 50 minutes?

2 answers:

3 0

S<span>tart with 6 cats kill 6 rats in 6 minutes, so it takes 6 minutes for 1 cat to kill 1 rat
1 cat kills 1 rat in 6 minutes
now extend this rate to 50 minutes (* 8 1/3)
1 cat kills 8 1/3 rats in 50 minutes and finally multiply by 12 to get the number of cats necessary to kill 100 rats in 50 minutes
12 cats kill 100 in 50 minutes
if 6 cats kill 6 rats in 6 minutes, then
12 cats kill 12 rats in 6 minutes << double the number of cats>>
now multiply time by 8 1/3 to get to 50 minutes, and therefore the number of rats killed multiplied by 8 1/3
yep, these 12 cats will kill 100 rats in 50 minutes</span>

EastWind [94]4 years ago

3 0

It will take 12 rats to kill 100 rats in 50 minutes

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Can someone explain to me what this means

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Answer:

2*2*2*3*5

Step-by-step explanation:

It is saying that 120 is the same as 24*5, and that 24* 5 is the same as 5 *2*12, which is that same as 5*2*3*4, which is then the same as 5*2*3*2*2.

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3 years ago

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leonid [27]

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Step-by-step explanation:

Sum of all angles of triangle = 180

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3 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

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

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

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

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

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

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

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

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

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

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3 years ago

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Regression analysis involving one dependent variable and more than one independent variable is known as a. linear regression. b.

Greeley [361]

Answer:

<h3>The Option B) Multiple regression is correct</h3><h3>Regression analysis involving one dependent variable and more than one independent variable is known as <u>multiple regression.</u></h3>

Step-by-step explanation:

Given that regression analysis involving one dependent variable and more than one independent variable

For : Regression analysis involving one dependent variable and more than one independent variable is known as <u>multiple regression</u>

In statistics, a linear regression is a linear relationship between a dependent variable and one or more independent variables.

In statistics for more than one independent variable and with one dependent variable in regression analysis , then the regression is called as multiple regression.

Therefore Option B) Multiple regression is correct

<h3>Regression analysis involving one dependent variable and more than one independent variable is known as <u>multiple regression</u>.</h3>

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