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klasskru [66]
3 years ago
8

Which function models the area of a rectangle with side lengths of 2x - 4 units x - 1 units

Mathematics
1 answer:
elena-s [515]3 years ago
6 0

The choices are missing but we can find the answer from the given information

Answer:

The function which models the area of the rectangle is A(x) = 2x² - 6x + 4

Step-by-step explanation:

The formula of the area of a rectangle is A = l w, where l is its length and w is its width

∵ The length of a rectangle is (2x - 4) units

∵ The width of the rectangle is (x - 1) units

- Use the formula of area to find its area

∵ A = l w

∴ A = (2x - 4)(x - 1)

Let us multiply the two brackets

∵ (2x - 4)(x - 1) = (2x)(x) + (2x)(-1) + (-4)(x) + (-4)(-1)

∴ (2x - 4)(x - 1) = 2x² + (-2x) + (-4x) + 4

- Add the like terms

∴ (2x - 4)(x - 1) = 2x² + (-6x) + 4

∴ (2x - 4)(x - 1) = 2x² - 6x + 4

Write A as a function of x

∵ The area of the rectangle is A(x)

∵ The area of the rectangle is 2x² - 6x + 4

∴ A(x) = 2x² - 6x + 4

The function which models the area of the rectangle is A(x) = 2x² - 6x + 4

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abruzzese [7]

Answer:

The answer is <em>1</em>.

Step-by-step explanation:

Given the expression:

|z-6|-|z-5|,\ if\ z

To find:

The expression without absolute value.

Solution:

First of all, let us learn about the absolute value function:

y = f(x) = |x| =\left \{ {{x\ if\ x>0} \atop {-x\ if\ x

i.e. value is x if x is positive

value is -x if x is negative

Here the given expression contains two absolute value functions:

|z-6| and |z-5|

Using the definition of absolute value function as per above definition.

|z-5| =\left \{ {{(z-5)\ if\ z>5} \atop {-(z-5)\ if\ z

|z-6| =\left \{ {{(z-6)\ if\ z>6} \atop {-(z-6)\ if\ z

Now, it is given that z < 5 that means z will also be lesser than 6 i.e. z < 6

So, given expression |z-6|-|z-5|,\ if\ z will be equivalent to :

-(z-6) - (-(z-5))\\\Rightarrow -z+6 +z-5 = \bold{1}

So, the expression is equivalent to <em>1</em>.

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Step-by-step explanation:

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Answer:

The mean is \mu = 131

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

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What is the mean of this normal distribution if the probability of scoring above x = 209 is 0.0228?

This means that when X = 209, Z has a pvalue of 1-0.0228 = 0.9772. So when X = 209, Z = 2.

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The mean is \mu = 131

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2 years ago
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