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torisob [31]
3 years ago
11

What is the solution to the system of equations? y = x +2 y = 3x -2

Mathematics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

x=2 and y= 4

Step-by-step explanation:

y = x +2 ----------(i)

y = 3x -2--------(ii)

Now, comparing (i) and (ii)

x+2= 3x-2

=>x-3x = -2-2

=>-2x= -4

=>2x= 4

=>x= 2

From eq (i)

y= x+2 = 2+2= 4

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The length of a rectangle is 7 inches longer than it is wide. If the area is 20 square inches, what are the dimensions of the re
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Area of a rectangle is length x width.

Let the width = x

The length would be x +7. ( 7 inches longer than the width)

Area = 20

Set up the formula:

20 = x * x+7

Simplify the right side:

20 = x^2 + 7x

Subtract 29 from both sides:

X^2 + 7x -20 = 0

Solve using the quadratic equation

X = -b + sqrt(b^2 -4ac) / 2a

X = -7 + sqrt(7^2-4(1)(-20) / 2(1) (exact answer)

X = 2.178908 ( as a decimal)

The width is 2.178907 inches (round as needed)

The length would be 9.178907 inches ( round as needed.)

Depending on how you round, when you multiply them together you get approximately 20 square inches.

4 0
3 years ago
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A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?
nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

                P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean SAT math score for in-state applicants = 540

\bar X_2 = sample mean SAT math score for out-of-state applicants = 555

s_1 = sample standard deviation for in-state applicants = 20

s_2 = sample standard deviation for out-of-state applicants = 25

n_1 = sample of in-state applicants = 35

n_2 = sample of out-of-state applicants = 35

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }  = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means (\mu_1-\mu_2) is ;

P(-1.997 < t_6_8 < 1.997) = 0.95  {As the critical value of t at 68 degree

                                         of freedom are -1.997 & 1.997 with P = 2.5%}  

P(-1.997 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.997) = 0.95

P( -1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

P( (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

<u>95% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

=[(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } },(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

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