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DaniilM [7]
3 years ago
9

Lindsay is 5 55 years younger than Mark. Seven years ago, the sum of their ages was 31 3131. Let l ll be Lindsay's age and let m

mm be Mark's age. Which system of equations represents this situation?
Mathematics
1 answer:
aev [14]3 years ago
7 0

Answer:

The system of equations that represents this situation is

M - L = 55

L + M = 45

Step-by-step explanation:

Let age of Lindsay is  = L

Age of mark = M

Given that

Lindsay is  55 years younger than Mark.

⇒ L = M - 55

⇒ M - L = 55 ------ (1)

Seven years ago, the sum of their ages was 31.

⇒ ( L - 7 ) + (M - 7) = 31

⇒ L + M = 45 ------- (2)

Therefore the system of equations that represents this situation is

M - L = 55

L + M = 45

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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} } ]

                                                 = [19.891 , 24.909]

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

6 0
3 years ago
Explain weather 14t-3t is equivalent to 3t-14t. Support your sender by evaluating the expression for t=2.
m_a_m_a [10]

Answer:

Solving 14t-3t we get 22 and 3t-14t we get -22

So, 14t-3t is not equivalent to 3t-14t.

Step-by-step explanation:

We need to explain weather 14t-3t is equivalent to 3t-14t. Support your sender by evaluating the expression for t=2.

<em>Equivalent expressions are those that have same values for any value of variable substituted.</em>

Now, We check if our expressions are equivalent, by evaluating the expression for t=2.

If they are equivalent, they would have same result after evaluation.

First, put t=2 into 14t-3t

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Now put t=2, into 3t-14t

3t-14t\\Put\:t=2\\=3(2)-14(2)\\=6-28\\=-22\\

For solving 14t-3t we get 22 and 3t-14t we get -22

So, 14t-3t is not equivalent to 3t-14t.

6 0
3 years ago
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Sergeeva-Olga [200]

9514 1404 393

Explanation:

This is a self-answering question: you solve it by graphing the equations.

<em>The solution is where the lines intersect</em>. The point of intersection of the lines is the point that satisfies all the equations for the lines, hence is a solution to the system. If they do not intersect, there are no solutions. If the lines are coincident, there are an infinite number of solutions.

__

The equations can be graphed by any of a number of methods. (My favorite is to let a graphing calculator do it.) The method of choice depends on the coefficients and the form the equations are given in. Methods of graphing are a topic for a more lengthy discussion.

6 0
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Answer:

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3 years ago
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Answer:

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8 0
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