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klemol [59]
3 years ago
6

Solve each equation. 0.2x2 + 1.2 = 3.4

Mathematics
2 answers:
Marat540 [252]3 years ago
6 0
0.2x^2+1.2=3.4\ \ \ |-1.2\\\\0.2x^2=2.2\ \ \ |:0.2\\\\x^2=11\to x=\pm\sqrt{11}\\\\\text{Answer:}\ x=-\sqrt{11}\ \vee\ x=\sqrt{11}
Irina-Kira [14]3 years ago
6 0

Answer:

x=-\sqrt{11}\text{ or }x=\sqrt{11}

Step-by-step explanation:

We have been given an equation 0.2x^2+1.2=3.4. We are asked to solve the given equation.

0.2x^2+1.2-1.2=3.4-1.2

0.2x^2=2.2

\frac{0.2x^2}{0.2}=\frac{2.2}{0.2}

x^2=11

Take square root of both sides:

x=\pm\sqrt{11}

x=-\sqrt{11}\text{ or }x=\sqrt{11}

Therefore, the solutions for our given equation would be x=-\sqrt{11}\text{ or }x=\sqrt{11}.

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AleksAgata [21]

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2 x 2.49 =

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5 0
3 years ago
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Gennadij [26K]

Answer:

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8 0
2 years ago
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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} } ]

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

6 0
3 years ago
A bag contains 10 red marbles, 15 yellow marbles, 5 green marbles, and 20 blue marbles. Two marbles are drawn from the bag.
kherson [118]

Answer:

It is the third option: (10C1) (20C1( / 50C2.

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6 0
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Step-by-step explanation:

<u>Step 1:  Distribute</u>

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5 0
3 years ago
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