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2 years ago

If

Mathematics

1 answer:

Elena L [17]2 years ago

3 0

Evaluate the derivative of f(x) at x = 1. I'll use the limit definition:

$\displaystyle \lim_{x\to1} \frac{f(x)-f(1)}{x-1} = \lim_{x\to1} \frac{6x^2-x^3-5}{x-1} = \lim_{x\to1}(-x^2+5x+5) = 9$

Then using the point-slope formula, the equation of the tangent line to f(x) at the point (1, 5) is

y - 5 = 9 (x - 1)

y = 9x - 4

You might be interested in

Which of the following comparisons is correct?<br>-2 2<br>4<−4<br>−6< −5

zlopas [31]

<h3>Answer: Choice D) -6 < -5</h3>

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Explanation:

Let's go through the answer choices

A) This is false because -2 is actually larger than -5. Make a number line and you'll see that -2 is to the right of -5. Numbers on the right are larger than numbers on the left.
B) This is also false. The value 0 is smaller than 2. A number line might help show this.
C) Also false. It should be 4 > -4. Any positive number is always larger than any negative number.
D) This is true. On a number line, -6 is to the left of -5, so -6 is smaller than -5. This makes -6 < -5 a true statement.

4 0

3 years ago

Assume y=x²+3x+6. what is the avg rate of change of y with respect to x as x changes from -4 to -2

AURORKA [14]

The average rate of change is 3

5 0

3 years ago

Give a direct proof that the sum an even integer and an odd integer is odd.

mina [271]

Even integer: 2 x a, where a is a natural number without 0;
Odd integer : 2 x b + 1 , where b is a natural number;
So, 2 x a + 2 x b+ 1 = 2 x ( a + b ) + 1, which is an odd integer.

5 0

3 years ago

Solve for x.<br> -6x = -18

marta [7]

Answer:

Step-by-step explanation:

The goal is to get the x to be on one side of the equation on its own.

-6x = -18

To get rid of the -6, we would have to do the opposite of what it's doing in the equation - in this case, we would divide, as it's the opposite of multiplying.

What we do to one side of the equation, we have to do to the other.

-6x/-6 = x

-18/-6 = 3

x = 3

Hope that makes sense!

3 0

3 years ago

Read 2 more answers

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago