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dimulka [17.4K]
3 years ago
12

In a standard normal distribution 95% of the data is within +/- _______standard deviations of the mean.

Mathematics
2 answers:
raketka [301]3 years ago
8 0
Your answer is 2 okay
dybincka [34]3 years ago
6 0

Answer:

2

Step-by-step explanation:

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Here are the first four terms in an arithmetic sequence.
finlep [7]

Answer:

88

Step By Step

Step-by-step explanation:

An arithmetic sequence is defined as

an=a+d(n-1) where an=nth term, a=initial term, n=term number, and d=common difference

For sequence 1,4,7... a=1 and d=4-1=7-4=3 so

an=1+3(n-1)

an=3n-2, so the 30th term is

an=3(30)-2

an=88

7 0
3 years ago
Read 2 more answers
Anybody? need help asap
AfilCa [17]
The answer would x=-1 for this question

6 0
3 years ago
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Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
andriy [413]

Answer:

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Step-by-step explanation:

We have been given a indefinite integral \int \frac{2}{3x\left(3x-5\right)}dx. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}

\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx

\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx

Using difference rule of integrals, we will get:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)

Now, we need to use u-substitution as:

Let u=3x-5.

\frac{du}{dx}=3

dx=\frac{1}{3}du

\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|

\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|

Substitute back these values:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)

Let us add a constant C.

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Therefore, our required integral would be \frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C.

5 0
3 years ago
The product of two numbers multiplied by 7
nadezda [96]

Answer:

7(x times y)

Step-by-step explanation:

Product means multiply, so it wants you to multiply two numbers and multiply THAT by 7.  So we will use two variables, in this case, I will use x and y.  X times y.  7 Times that.

7 0
3 years ago
Read 2 more answers
helene invested a total of $1000 in two simple-interest bank accounts. one account paid 5% annual interest; the other paid 6% an
alisha [4.7K]
Helene invested $200 for the 5% annual interest rate and $800 for the 6% annual interest rate.
8 0
3 years ago
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