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3 years ago

In an experiment, the one factor that the scientist changes is called what?

Mathematics

2 answers:

IrinaK [193]3 years ago

7 0

Answer:

independent

Step-by-step explanation:

kotegsom [21]3 years ago

5 0

Answer:

Independent Variable

Step-by-step explanation:

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Please help ! Math -

harkovskaia [24]

Answer:

f(x) = 2(x - 2)^2 - 1

vertex: (2, -1)

Step-by-step explanation:

f(x) = 2x^2 - 8x + 7

First, we find the vertex.

x = -(-8)/4 = 2

y = 2(2)^2 - 8(2) + 7 = 2(4) - 16 + 7 = 8 - 9 = -1

vertex: (2, -1)

Second, we write f(x) in vertex form.

we know that h and k have to be 2 and -1.

2x^2 - 8x + 7 = a(x - 2)^2 - 1

Since 8 - 1 = 7, we do this:

f(x) = (2x^2 - 8x + 8) - 1

Factor out the 2 and then factor the polynomial

f(x) = 2(x^2 - 4x + 4) - 1

factors of 4:

1 4

2 2

-1 -4

-2 -2 = -4

the function in vertex for is:

f(x) = 2(x - 2)^2 - 1

8 0

3 years ago

A figure is made from 8 units blocks.Its 3 units tall.What is the maximum length the figure could be?Explain.

Aleksandr-060686 [28]

i think the maximum length of the figure would be : 6 The total blocks that exist to form should be 8 we know that 3 is used for the height , so the remaining will be 5. but we can use the first blocks as the foundation for the additional length. The figure would look like this : O O OOOOOO ---------> total of 8 blocks

6 0

3 years ago

Apply the distributive property and simplify -6(x+2)=36

Delicious77 [7]

Answer:

-8

Step-by-step explanation:

-6(x+2)=36

-6x-12=36

-6x=48

x=-8

5 0

3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

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

In which

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

$\mu = \lambda = 4(3) = 12$

$\sigma = \sqrt{\lambda} = \sqrt{12}$

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 12}{\sqrt{12}}$

$Z = 2.31$

$Z = 2.31$ has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0

3 years ago

Geometry matching help

o-na [289]

Answer:

Ellipse C

Circle E

Sphere B

Great Circle D

Triangle A

Step-by-step explanation:

Cross sections are slices through a 3d figure to create a 2d shape. A cone creates a circle and triangle. A sphere creates a great circle. A cylinder creates an ellipse. Rotating a semi circle around an axis creates a sphere.

7 0

3 years ago