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3 years ago

AP calculus HW Need help on #61

Mathematics

1 answer:

Yuki888 [10]3 years ago

7 0

Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/∆x

Crossing the 1s , the 2xs, and the x^2s out, we get

(2x∆x+∆x^2-2∆x)/∆x

Dividing the ∆x, we get
2x+∆x-2.

Making the limit of ∆x=0, we get 2x-2.

You might be interested in

What is the solution to the system? (___,___)

Rina8888 [55]

For the first line we have a slope of (y2-y1)/(x2-x1)

(2--2)/(1--1)=4/2=2 so we have:

y=2x+b, now solve for b with either of the points, I'll use: (1,2)

2=2(1)+b

b=0 so the first line is:

y=2x

Now the second line:

(1-10)/(4--2)=-9/6=-3/2 so far then we have:

y=-3x/2+b, using point (4,1) we solve for b...

1=-3(4)/2+b

1=-6+b

b=7 so

y=-3x/2+7 or more neatly...

y=(-3x+14)/2

...

The solution occurs when both the x and y coordinates for each are equal, so we can say y=y, and use our two line equations...

2x=(-3x+14)/2

4x=-3x+14

7x=14

x=2, and using y=2x we see that:

y=2(2)=4, so the solution occurs at the point:

(2,4)

6 0

3 years ago

Which shows how to solve the equation Three-fourths x = negative 6 for x in one step?

tino4ka555 [31]

Answer:

x = -8

Step-by-step explanation:

Let's put this into numbers: $\frac{3}{4}$ x =-6

The first thing you want to do is to multiply both sides by 4:

3x=-24

Next you need to divide by 3:

x=-8

The answer is -8

5 0

3 years ago

The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to in

oksano4ka [1.4K]

Answer:

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

$p_v =P(Z>2)=0.0228$

Step-by-step explanation:

1) Data given and notation

$\bar X=8300$ represent the sample mean

$\sigma=1200$ represent the population standard deviation

$n=64$ sample size

$\mu_o =800$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

4) P-value

Since is a one-side upper test the p value would given by:

$p_v =P(Z>2)=0.0228$

5) Conclusion

If we compare the p value and the significance level assumed, for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.