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aliya0001 [1]
3 years ago
11

Which expression is equivalent to 36a-12b? 3(6a-2b). 6(6a-2b). 3(6a+2b). 6(6a+2b)

Mathematics
2 answers:
exis [7]3 years ago
7 0

Answer:

6(6a-2b)

Step-by-step explanation:

if you multiply the 6 by what's in parenthesis you get 36a-12b

Maurinko [17]3 years ago
4 0

Answer:

6(6a-2b)

Step-by-step explanation:

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What is a logarithm? if your employer asked you to select one of these options: option 1: your salary for the next two months wi
Paladinen [302]
Take option 2.
Let n=2^30
Then:
ln n=ln 2^30=30 ln 2
ln n=20.8
n=e^20.8, or 1,073,741,824
1073741824 x.01=$10,737,418.24 due in salary at the end of the first month!!!!!
8 0
3 years ago
Who is the general information of acounting
Anettt [7]

Answer:

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6 0
2 years ago
Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
What is the answer to 132.55=5 (0.04m+21.70+0.10(21.70))
postnew [5]
132.55=5(.04m+2+.7+0.1(21.7))>>>m=66
8 0
3 years ago
What are the answers
Shalnov [3]
Slope formula: 
y2-y1
-------
x2-x1
    48)
1-3       -2
----    = ----
-2-4     -6

So the slope of MN is 1/3
5 0
3 years ago
Read 2 more answers
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