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3 years ago

Simplify your answer as much as possible

Mathematics

1 answer:

loris [4]3 years ago

7 0

9514 1404 393

Answer:

-4u^2/y+(5/2)y^3u^2

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

$(8y^2u^7-5y^6u^7)\div(-2y^3u^5)=\dfrac{8y^2u^7-5y^6u^7}{-2y^3u^5}\\\\=-4y^{2-3}u^{7-5}+\dfrac{5}{2}y^{6-3}u^{7-5}=\boxed{\dfrac{-4u^2}{y}+\dfrac{5}{2}y^3u^2}$

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Is this statement true or false?

bekas [8.4K]

Answer:

A:False

Step-by-step explanation:

4 0

3 years ago

At Burnt Mesa Pueblo, archaeological studies have used the method of tree-ring dating in an effort to determine when prehistoric

Marina CMI [18]

Answer:

a) The range is (1199, 1267)

b) The range is (1165, 1301)

c) The range is (1131, 1335)

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(1233,34)$

Where $\mu=1233$ and $\sigma=34$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

Part a

For this case we can use the statement from the empirical rule "68% of the data falls within the first standard deviation (µ ± σ)", and we can find the limits like this:

$\mu -\sigma= 1233-34=1199$

$\mu +\sigma=1233+34=1267$

The range is (1199, 1267)

Part b

For this case we can use the statement from the empirical rule "95% of the data within the first two standard deviations (µ ± 2σ)", and we can find the limits like this:

$\mu -2\sigma= 1233-(2*34)=1165$

$\mu +2\sigma=1233+(2*34)=1301$

The range is (1165, 1301)

Part c

For this case we can use the statement from the empirical rule "99.7% of the data within the first three standard deviations (µ ± 3σ)" and that represent almost all the data, and we can find the limits like this:

$\mu -3\sigma= 1233-(3*34)=1131$

$\mu +3\sigma=1233+(3*34)=1335$

The range is (1131, 1335)

5 0

3 years ago

Integrals using U sub. Please show steps so I can learn too :)

Romashka [77]

Answer:

$\frac{arctan^2(x)}{2}$

Step-by-step explanation:

For this question, set u = arctan(x). This would be the easiest way because the derivative of arctan(x) is $\frac{1}{x^2+1}$ which is what we have. Before setting u, always look at the question and think about possible derivatives.

$\int\frac{arctanx}{x^2+1}dx \ Let \ u = arctan(x)\\\\\\u = arctan(x) \ \ \ du = \frac{1}{x^2+1}dx\\$

Next, plug in u and du.

$\int (arctan(x)*\frac{1}{x^2+1})dx$ (rearranging)

$\int u\ du$

$= \frac{u^2}{2} = \frac{1}{2}u^2$

Substitute arctan(x) back in for u and add +C.

$\frac{arctan^2(x)}{2}+C$

7 0

3 years ago

a girl has to be at school at 8:00 am it takes her 20 minutes to eat, 35 minutes to get dressed and 15 minutes to get dressed wh

mixer [17]

First find the sum of the amount of time it takes her to be finished

20 + 35 + 15 =
35 + 35 = 70

Keep in mind a hour is 60 minutes

1:10 minutes before 8:00

She should get up at 6:50am

4 0

3 years ago

3x2 − y3 − y3 − z if x = 3, y = −2, and z = −5.

allsm [11]

The correct answer is 35. Your Welcome!

7 0

3 years ago