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3 years ago

What’s the slope I need help!

Mathematics

2 answers:

Grace [21]3 years ago

4 0

Answer:

- $\frac{1\\}{3}$

idk if its right. make sure to tell me :)

Lilit [14]3 years ago

3 0

Answer:

-2\7 I think

Step-by-step explanation:

mafffffffffgfgggggfffs

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WHAT IS 1+100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000=

Keith_Richards [23]

Answer:

100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,001

Step-by-step explanation:

Hope this helps!:)

7 0

3 years ago

Read 2 more answers

Suppose a motorcycle costs $8,000 and loses 4% of its value each year. What will be the value of the motorcycle after 7 years?

Mazyrski [523]

Answer:the answer is 6,012 after 7 years

Step-by-step explanation:

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3 years ago

A survey is to be conducted to determine the average driving in miles by Minnesota State University, Mankato students. The inves

Nadusha1986 [10]

Answer:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =1.5 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04;0;1)", and we got $z_{\alpha/2}=1.75$ , replacing into formula (b) we got:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 rounded up to the nearest integer

5 0

3 years ago

The graph of F(x), shown below, has the same shape as the graph of G(x) = x2, but it is shifted up 3 units. What is its equation

Aliun [14]

For this case, the parent function is given by:
$G (x) = x ^ 2
$
Applying the following transformation we have:
Vertical displacement
Assume k> 0,
To graph y = f (x) + k, move the graph k units up.
We have then:
$F (x) = G (x) + 3

F (x) = x ^ 2 + 3$
Answer:
the equation of F (x) is given by:
$F (x) = x ^ 2 + 3$

5 0

3 years ago

Please help with this Calculus questions

Triss [41]

Answer:

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93$ .

Step-by-step explanation:

To find $\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du$ .

First, calculate the corresponding indefinite integral:

Integrate term by term:

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant rule $\int c\, du = c u$

$\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$

$2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$

$2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}$

Therefore,

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)$

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$ , so just evaluate the integral at the endpoints, and that's the answer.

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}$

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0$

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}$

6 0

3 years ago