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3 years ago

Find f(-3) if f(x) = x 2

Mathematics

1 answer:

kobusy [5.1K]3 years ago

8 0

Answer:

The value of given function for x = - 3 is 9

Step-by-step explanation:

Given as :

The y is the function of x is as

f( x ) = x²

So, For any value of x , the value of f( x ) should change

If, x = - 3

So, f ( - 3 ) = ( - 3 )²

Or, f ( - 3 ) = ( - 3 ) × ( - 3 )

i.e f ( - 3 ) = ( - × - ) ( 3 × 3 )

Or, f ( - 3 ) = ( + ) ( 9 )

∴ f ( - 3 ) = 9

So, The value of function f ( - 3 ) = 9

Hence , The value of given function for x = - 3 is 9 Answer

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The length of the blue ribbon is two-thirds the length of the red ribbon. Write an equation to find the length r of the red ribb

vladimir2022 [97]

The equation to find the length r of the red ribbon is:

$r = \frac{3b}{2}$

Solution:

Let "b" be the length of blue ribbon

Let "r" be the length of red ribbon

To find: equation for length r of the red ribbon.

The length of the blue ribbon is two - thirds the length of the red ribbon

Which means,

$\text{Length of blue ribbon } = \frac{2}{3} \times \text{ length of the red ribbon }$

Therefore,

$b = \frac{2}{3} \times r\\\\3b = 2r\\\\r = \frac{3b}{2}$

Thus the equation to find the length r of the red ribbon is found

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3 years ago

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Can someone please help me with this problem

alukav5142 [94]

Answer:

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Step-by-step explanation:

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4 years ago

Problem 2.9 a runner completes the 200-meter dash with a time of 20.19 seconds. part a what was the runner's average speed in me

Elina [12.6K]

Answer:

Average speed = 9.91 m/s

Explanation :

Total distance covered by a runner = 200 m

Time taken to complete this distance = 20.19 sec

We need to find the average of runner speed in meter/sec

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$speed = \frac{distance}{time}$

speed = $\frac{200}{20.19}$

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that's the final answer.

I hope it will help you.

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3 years ago

⚠️⚠️24 points⚠️⚠️

motikmotik

Answer: No. Equations even if it is paired. So on. It can only have one solution. Although this is only for a system of equations.

A linear can have two. It just depends on what the problem is.

Step-by-step explanation:

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3 years ago

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A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their

Sever21 [200]

Answer:

A. Normal

B. Between 40.08 minutes and 43.92 minutes.

C. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

Step-by-step explanation:

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, a normal distribution.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{12}{\sqrt{150}} = 1.92$

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

Between 40.08 minutes and 43.92 minutes.

Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

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