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3 years ago

Choose the equation that represents a line that passes through points (−3, 2) and (2, 1).

Mathematics

1 answer:

guapka [62]3 years ago

8 0

Answer:

$y=-\frac{1}{5}x+\frac{7}{5}$

Step-by-step explanation:

The given line passes through points (−3, 2) and (2, 1).

We use the formula $m=\frac{y_2-y_1}{x_2-x_1}$ to find the slope of this line.

$\implies \frac{1-2}{2--3}=\frac{-1}{5}$

We use the point slope formula:

$y-y_1=m(x-x_1)$

$y-2=-\frac{1}{5}(x--3)$

$y-2=-\frac{1}{5}(x+3)$

We expand to get:

$y-2=-\frac{1}{5}x-\frac{3}{5}$

$y=-\frac{1}{5}x-\frac{3}{5}+2$

The slope intercept form of the line is:

$y=-\frac{1}{5}x+\frac{7}{5}$

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A jar of peanut butter contains 454 g with a standard deviation of 10.2 g. Find the probability that a jar contains more than 46

Fofino [41]

Answer:

The probability that a jar contains more than 466 g is 0.119.

Step-by-step explanation:

We are given that a jar of peanut butter contains a mean of 454 g with a standard deviation of 10.2 g.

Let X = Amount of peanut butter in a jar

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 454 g

$\sigma$ = standard deviation = 10.2 g

So, X ~ Normal( $\mu=454 , \sigma^{2} = 10.2^{2}$ )

Now, the probability that a jar contains more than 466 g is given by = P(X > 466 g)

P(X > 466 g) = P( $\frac{X-\mu}{\sigma}$ > $\frac{466-454}{10.2}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.881 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

4 0

3 years ago

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

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A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

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The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.