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vaieri [72.5K]
3 years ago
9

Convert 25 miles into kilometres

Mathematics
1 answer:
Airida [17]3 years ago
3 0

Answer:

1 miles= 1.609km

so, 25x1.609 = 40.225km

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Write an equation in slop-intercept form the line with slope 1/4 and y-intercept -7
Licemer1 [7]

When you write an equation in the slope-intercept form

y=mx+b

the two coefficients m and b represent, respectively, the slope and the intercept.

So, in your case, m=1/4 and b= -7, which leads to the equation

y=\dfrac{1}{4}x-7

7 0
4 years ago
20 points!! help needed
dimaraw [331]

ANSWER

1 bell shape

2 to find probability when sampling

EXPLANATION

1 In a normal distribution, the mode,mean and median are equal.

As a result, the distribution is neither skewed to the right or left.

The shape of the normal distribution looks like a bell.

That is why it is also called the bell curve.

2. The area under the normal curve is 1.

The line of symmetry of the bell shaped distribution divides it into two halves with area 0.5 each.

The normal curve is therefore used to find the probabilities of a sample distributions.

7 0
3 years ago
What is this equation simplified<br><img src="https://tex.z-dn.net/?f=%20%28%7Bx%7D%5E%7B3%7D%292" id="TexFormula1" title=" ({x}
SVEN [57.7K]

Answer:

2x³

Step-by-step explanation:

2x³

5 0
3 years ago
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PtichkaEL [24]

Answer:

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3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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