I need help on this equation, it’s solutions to inequalities. I need explanation on why it’s that answer, plz, thx

If g(x)=a x f(x+b) how is f(x) transformed to get g(x)

guajiro [1.7K]

Answer:

Ok, we know that g(x) = a*f(x + b), we want to find all the transformations that we must apply to f(x) to construct g(x).

Let's assume at the beggining that T2(T1(f(x))) = g(x), and start appliying the transformations.

First, we have a horizontal shift.

If we want to move the graph A (A positive) units to the right, we should have:

g(x) = f(x - A)

in this case we have

g(x) = f(x + b) = f(x - (-b))

So we have a shift of -b units to the right, or b units to the left.

Second:

We have a vertical contraction/dilation.

if we have a function f(x), a compression/dilation of scale factor A is written as:

g(x) = A*f(x)

if A > 1, we have a dilation.

if 0 < A < 1, we have a contraction.

In this case the scale factor is a.

g(x) = a*f(x).

Now, applying both transformations we have:

> shift of b units to the left.

> contraction/dilation of scale factor a.

g(x) = a*f(x +b)

4 0

2 years ago

Read 2 more answers

Whats the answer to 8.4 + 0.18x = 0.32x

Sveta_85 [38]

Step-by-step explanation:

$0.32x - 0.18x = 8.4 \\ 0.14x = 8.4 \\ x = \frac{8.4}{0.14} \\ x = 60$

3 0

1 year ago

The table shows the prices of 8 different treadmills at a sporting goods

irakobra [83]

Answer:

mean: $837.50; median: $810; mode: $810; range: $645

Step-by-step explanation:

Discounted prices are given as:

$640, $810, $755, $810, $1175, $900, $1080, $530

a. Mean = Sum of the Number of Term ÷ Total number of Terms

Total number of term = 8

Mean = ($640 + $810 + $755 + $810 + $1175 + $900 + $1080 + $530) ÷ 8

Mean = $837.50

b) Median.

$640, $810, $755, $810, $1175, $900, $1080, $530

The first step is to rearrange the discounted prices =

$530, $640, $755, $810, $810, $900, $1080, $1175

Second step is to find the number in the middle

$530, $640, $755, )$810, $810, ($900, $1080, $1175

= ($810 + $810) ÷ 2

= $1920 ÷ 2

= $810

Median = $810

c) Mode

This is calculated as the prices that occurs the most or more than once amongst the Discounted prices.

Discounted prices = $530, $640, $755, $810, $810, $900, $1080, $1175

Price that occurs more than once = $810.

Mode = $810

d. Range

This is calculate as the Highest Discounted price - Lowest Discounted price.

Highest Discounted price = $1175

Lowest Discounted price =$530

Range = $1175 - $530

= $645

6 0

2 years ago

Find the mean, median, and mode of the data set. Round to the nearest tenth. test scores on a math exam:

irinina [24]

Answer:

Mean: $\frac{89+93+76+89+68+80+89+83+88+87+63+86+73+74+67+93+68+95+66+99+78+100 }{22}$

Median: 63, 66, 67, 68, 73, 74, 76, 78, 80, 83, 86, 87, 88, 89, 89, 89, 93, 93, 95, 99, 100

$\frac{86+83}{2}$ = 169/2

Mode: 100

Step-by-step explanation:

8 0

1 year ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2>0$

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

2 years ago