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jolli1 [7]
4 years ago
12

Find the value of the determinant.

Mathematics
1 answer:
anastassius [24]4 years ago
4 0

Answer:33

Step-by-step explanation: i just did the problem

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natima [27]
100203 = 1.00203 x 10^5
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3 years ago
We wish to construct a 90% confidence interval on the mean of a population known to be normally distributed based on a sample of
sergiy2304 [10]

Answer:

Margin of error = 0.773

Step-by-step explanation:

We are given the following information in the question.

Sample size, n = 20

Sample mean = 8

Sample standard deviation = 2

90% Confidence interval

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729  

Margin of error:

t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

1.729(\frac{2}{\sqrt{20}} ) = 0.773  

7 0
4 years ago
PLSS HELP ME THE QUESTION AND EVERYTHING IS ABOVE what is the slope of the line and y intercept
AURORKA [14]

Answer:

C

Step-by-step explanation:

It is C. Have a great day!

6 0
3 years ago
If you spend $150 on 3 items and 1 item cost 50% more than the other 2 How much does it cost?
ryzh [129]
$75 Hope that helps.

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{
il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
2 years ago
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