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Help!!!!!!!<br> The functions f(x) and g(x) are shown on the graph.<br> f(x) = x2<br> What is g(x)?

marysya [2.9K]

Answer:

-x²-2

Step-by-step explanation:

6 0

3 years ago

X2 - 2x + 2 = 0

seraphim [82]

Answer:

Step-by-step explanation:

Let's solve x^2 - 2x + 2 = 0 using "completing the square:"

1. Take the coefficient of x: It is -2.

2. Halve this, obtaining -1.

3. Square this result, obtaining 1.

4. Add 1, and then subtract 1, between -2x + 2:

x^2 - 2x + 1 - 1 + 2 = 0

5. Rewrite x^2 - 2x + 1 + 1 = 0 beginning with the square of a binomial

(x - 1)^2 + 1 = 0, or (x - 1)^2 = -1

6. Take the square root of both sides, obtaining x - 1 = ±i, or x = 1 ±i

7. Write out the roots: they are x = 1 + i and x = 1 - i (two complex, different roots). No real roots, so the last command of this question is irrelevant. The graph never touches the x-axis; the graph is in Quadrants I and II and is that of a parabola that opens up.

6 0

3 years ago

The scores of eight grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.

rewona [7]

Work out the z scores for 51 and 54
z1 = (51 - 57.5) / 6.5 = -1
z2 = (64 - 57.5)/6.5 = 1
from tables of normal distribution this value os 3413 + 3413 = 68.26%

so the answer is c

8 0

4 years ago

Since 1900, the magnitude of earthquakes that measure 0.1 or higher on the Richter Scale in CA are distributed normally with a m

Gwar [14]

Answer:

a) 3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b) 1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c) 73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d) 99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e) 6.0735

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 6.2, \sigma = 0.5$

a.) What is the probability that a randomly selected earthquake in CA has a magnitude greater than 7.1?

This is 1 subtracted by the pvalue of Z when X = 7.1. So

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

$Z = \frac{7.1 - 6.2}{0.5}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641

1 - 0.9641 = 0.0359

3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b.) What is the probability that a randomly selected earthquake in CA has a magnitude less than 5.1?

This is the pvalue of Z when X = 5.1. So

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

$Z = \frac{5.1 - 6.2}{0.5}$

$Z = -2.2$

$Z = -2.2$ has a pvalue of 0.0139

1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c.) What is the probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1?

Now $n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581$

This is 1 subtracted by the pvalue of when X = 6.1. So

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

By the Central Limit Theorem

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

$Z = \frac{6.1 - 6.2}{0.1581}$

$Z = -0.63$

$Z = -0.63$ has a pvalue of 0.2643

1 - 0.2643 = 0.7357

73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d.) What is the probability that a ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

This is the pvalue of Z when X = 7.22 subtracted by the pvalue of Z when X = 5.7. So

X = 7.22

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

$Z = \frac{7.22 - 6.2}{0.1581}$

$Z = 6.45$

$Z = 6.45$ has a pvalue of 1

X = 5.7

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

$Z = \frac{5.7 - 6.2}{0.1581}$

$Z = -3.16$

$Z = -3.16$ has a pvalue of 0.0008

1 - 0.0008 = 0.9992

99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e.) Determine the 40th percentile of the magnitude of earthquakes in CA.

This is X when Z has a pvalue of 0.4. So it is X when Z = -0.253.

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

$-0.253 = \frac{X - 6.2}{0.5}$

$X - 6.2 = -0.253*0.5$

$X = 6.0735$

6 0

3 years ago

Sharon is making a large batch of soup. The soup reaches a height of 25 in a cylindrical pot whose diameter is 30cm. To store th

nasty-shy [4]

Complete question :

Sharon is making a large batch of soup. The soup reaches a height of 25 in a cylindrical pot whose diameter is 30cm. To store the soup for later, she'll pour it into ice cube molds where each cube has edges that are 7cm long. How many whole cubes can Sharon make?

Answer:

About 51 ice cubes

Step-by-step explanation:

Given the following :

Height (h) of cylindrical pot = 25

Diameter = 30cm

Edge of ice cube = 7cm long

Volume of cylinder (v) = πr^2h

V = π * (30/2)^2 * 25

V = π * 15^2 * 25

V = 17671.458cm^3

Therefore, the soup occupies 17673.75cm^3

Volume of cubes to store the soup for later:

The volume of a cube is given by the formula:

V = a^3

Where a is the length of it's edge

V = 7^3

V = 343cm^3

Number of cubes required to store soup:

(Volume of cylinder / Volume of ice cube)

17671.458cm^3 / 343cm^3

= 51.520287

This is about 51 whole cubes

7 0

3 years ago