CAN SOMEBODY PLEASE HELP ME

Find all maxima and minima of the function given below. Use a horizontal span of -6 to 6. (Round your answers to two decimal pla

AVprozaik [17]

Equation: 4 + 4x + x²

Maxima: No maximum value from said equation
Minima: x = -2 since that is the only critical point in the equation

6 0

3 years ago

Does anyone know how to do this

Novay_Z [31]

Answer:

Six over Seven

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If -12-4(-2x)= 4 then what is x?

11111nata11111 [884]

Answer:

$\boxed{\boxed{\sf x=2}}$

Step-by-step explanation:

$\sf -12-4\left(-2x\right)=\:4$

First, let's remove parentheses:

$\sf -12-4\times -2x=4$

Simplify, multiply, 4 * -2x = -8x:

$\sf -12-(-8x)=4$

** Apply rule: - ( -a )= a

$\sf 12+8x=4$

Now, let's regroup terms

$\sf 8x-12=4$

Add 12 to both sides:

$\sf -12+8x+12=4+12$

Simplify, 4 + 12= 16

$\sf 8x=16$

Divide both sides by 8:

$\sf \cfrac{8x}{8}=\cfrac{16}{8}$

$\sf x=2$

6 0

3 years ago

Read 2 more answers

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

motikmotik

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$