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2 years ago

At a candy factory, butterscotch candies were

Mathematics

1 answer:

KIM [24]2 years ago

5 0

Answer:

the answer is 2

Step-by-step explanation:

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Given a dilation around the origin, what is the scale factor K?

kobusy [5.1K]

Answer:

The answer is " $\frac{4}{3}$ "

Step-by-step explanation:

Given:

$\bold{DoK = (12, 8) (9,6)}$

In the scale factor when we take x coordinate then the value of k is:

$\Rightarrow x= \frac{x_1}{x_2}\\\\\Rightarrow x= \frac{12}{9}\\\\\Rightarrow x= \frac{4}{3}\\$

If we take y coordinate then the value of k is:

$\Rightarrow y= \frac{y_1}{y_2}\\\\\Rightarrow y= \frac{8}{6}\\\\\Rightarrow y= \frac{4}{3}\\$

So, the final value of k is " $\frac{4}{3}$ " .

5 0

3 years ago

SOMEONE PLS HELP ME WITH THIS: Jack Thomas has an fudge sundae with 450 calories. How long would it take him to burn off the cal

KATRIN_1 [288]

Answer:

It would take him 450 minutes to burn off the calories by sleeping

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

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

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!

zmey [24]

Step-by-step explanation:

underneath the x axis is the minuses so you draw across from -4

4 0

3 years ago

Choose the constant term that completes the perfect square trinomial.<br><br> y2 + y

Oduvanchick [21]

Hello!

First off, please write y2 + y as y^2 + y. The " ^ " symbol denotes exponentiation, whereas y2 is meaningless.

To find the constant term in question, take half of the coefficient of y (that is, take 1/2) and square it. Then we have y2 + y + 1/4.

The constant term in question is 1/4.

3 0

3 years ago