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3 years ago

What is the answer to |3x-3|=9

Mathematics

1 answer:

BabaBlast [244]3 years ago

7 0

Answer:

x = 4, -2

Step-by-step explanation:

split the absolute value into 2 equations:

|3x - 3| = 9

3x - 3 = 9 3x - 3 = -9

3x = 12 3x = -6

x = 4 x = -2

x = 4, -2

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Find the arc length and area of a sector with radius R =60 inches and central angle equals 30° round your answer to the nearest

Black_prince [1.1K]

Answer:

(a)31.42 inches

(b)942.48 Square Inches

Step-by-step explanation:

Given a sector of a circle with the following dimensions:

Radius of the circle =60 inches

Central Angle of the sector =30°

(a)Arc Length

Arc Length $=\dfrac{\theta}{360^\circ}X2\pi r$

$=\dfrac{30}{360^\circ}X2*60*\pi\\\\=10\pi\\\\=31.42$ Inches (correct to the nearest hundredth)$

(b)Area of the sector

Area of the sector $=\dfrac{\theta}{360^\circ}X\pi r^2$

$=\dfrac{30}{360^\circ}X\pi*60^2\\\\=300\pi\\\\=942.48$ Square Inches (correct to the nearest hundredth)$

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3 years ago

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pav-90 [236]

Answer:

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Step-by-step explanation:

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3 years ago

Can anyone tell me why by direct substitution of x, the equation (circled ones) equals to the indeterminate form, 0/0? When you

Katena32 [7]

Answer:

See explanation and hopefully it answers your question.

Basically because the expression has a hole at x=3.

Step-by-step explanation:

Let h(x)=( x^2-k ) / ( hx-15 )

This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.

Solving for x in that equation:

Adding 15 on both sides:

hx=15

Dividing both sides by h:

x=15/h

For it be a hole, you also must have the numerator is zero at x=15/h.

x^2-k=0 at x=15/h gives:

(15/h)^2-k=0

225/h^2-k=0

k=225/h^2

So if we wanted to evaluate the following limit:

Lim x->15/h ( x^2-k ) / ( hx-15 )

Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.

We were ask to evaluate

Lim x->3 ( x^2-k ) / ( hx-15 )

Comparing the two limits h=5 and k=225/h^2=225/25=9.

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3 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

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

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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