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2 years ago

11

Explain what a, h, and k do to transform the following equation. Be sure to mention the shift or transformation each variable do

es and how if each varible is a positive or negative : y= a|x-h|+k.

2 answers:

dangina [55]2 years ago

3 0

Answer:

a: vertical stretch

h: horizontal translation (remember not to get signs mixed up!)

k: vertical translation

Lady_Fox [76]2 years ago

3 0

Explanation:

A → Whether opens down or up; fractions [vertical shrink], whole numbers [vertical stretch]

-H → OPPOSITE terms of what they really are [phase shift]

K → NORMAL term [vertical shift]

I am joyous to assist you anytime.

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There are 221 students on a field trip.

Liono4ka [1.6K]

The equation that models the students is a linear equation

The equation that models the number of students in each bus is 6x = 216, and the number of students in each bus is 36

<h3>How to determine the equation?</h3>

The given parameters are:

Students = 221

Van = 5

The rest students = 6 buses

Start by calculating the number of students remaining:

Remaining students = 221 - 5

Remaining students = 216

Represent the number of students in each bus with x.

So, we have:

6 buses * x = Remaining students

This gives

6x = 216

Divide both sides by 6

x = 36

Hence, the equation that models the number of students in each bus is 6x = 216, and the number of students in each bus is 36

Read more about linear equations at:

brainly.com/question/15602982

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1 year ago

Help help help math math ASAP

REY [17]

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2 years ago

Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.

____ [38]

Let $z=-\sqrt3+i$ . Then

$|z|=\sqrt{(-\sqrt3)^2+1^2}=2$

$z$ lies in the second quadrant, so

$\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6$

So we have

$z=2e^{i5\pi/6}$

and the fourth roots of $z$ are

$2^{1/4}e^{i(5\pi/6+k\pi)/4}$

where $k\in\{0,1,2,3\}$ . In particular, they are

$2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}$

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$2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}$

$2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}$

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2 years ago

Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636, the central limit theorem appl

lakkis [162]

Answer:

a) $\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1$

b) $P(\bar X >101)=1-P(\bar X$

c) $P(\bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=100,\sigma=6)$

And let $\bar X$ represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

a. What are the mean and variance of the sampling distribution for the sample means?

$\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1.2^2=1.44$

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833$

And we want to find this probability:

$P(\bar X >101)=1-P(\bar X$

On this last step we use the complement rule.

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

$z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67$

And we want to find this probability:

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3 years ago

What is the remainder for the synthetic division problem below?

Luden [163]

7 |1 -8 10
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1 -1 3

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1 year ago