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2 years ago

For each part below, give an example of a linear system of

Mathematics

1 answer:

N76 [4]2 years ago

7 0

Answer:

a)g: 3x + 4y = 10 b) a:x+y = 5 c) c: 3x + 4y = 10

h: 6x + 8y = 5 b:2x + 3y = 8 d: 6x + 8y = 5

Step-by-step explanation:

a) Has no solution

g: 3x + 4y = 10

h: 6x + 8y = 5

Above Equations gives you parallel lines refer attachment

b) has exactly one solution

a:x+y = 5

b:2x + 3y = 8

Above Equations gives you intersecting lines refer attachment

c) has infinitely many solutions

c: 3x + 4y = 10

d: 6x + 8y = 5

Above Equations gives you collinear lines refer attachment

i) if we add x + 2y = 1 to equation x + y = 5 to make an inconsistent system.

ii) if we add x + 2y = 3 to equation x + y = 5 to create infinitely system.

iii) if we add x + 4y = 1 to equation x + y = 5 to create infinitely system.

iv) if we add to x + y =5 equation x + y = 5 to change the unique solution you had to a different unique solution

You might be interested in

Match each point label on the box it to its description.

Ahat [919]

A is the minimum value
B is the first quartile
The point with the line through it (next to C) is the median.
The point to the outside of the box is the third quartile.
D is the maximum value.

Hope this helped.

4 0

2 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

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

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

The vertices of a polygon are P(0,4), Q(5,4), R(5,−3) and S(0,−3). What is

andrezito [222]

Answer:

Perimeter of the polygon p = 24

Step-by-step explanation:

Step(i):-

Given the vertices of a polygon are

P(0,4) ,Q( 5,4) ,R( 5,-3) and S(0,-3)

The distance of PQ

$a = PQ = \sqrt{(4-4)^{2} +(5-0)^{2} } = \sqrt{25} =5$

The distance of QR

$b = Q R= \sqrt{(-3-4)^{2} +(5-5)^{2} } = \sqrt{49} =7$

The distance of RS

$c = RS = \sqrt{(0-5)^{2} +(-3+3)^{2} } = \sqrt{25} =5$

The distance of PS

$d = PS = \sqrt{(0-0)^{2} +(-3-4)^{2} } = \sqrt{49} =7$

Step(ii):-

Perimeter of the polygon

= sum of all sides of polygon

p = a+ b+ c+ d

p = 5+7+5+7

p = 24

Final answer:-

Perimeter of the polygon p = 24

5 0

3 years ago

Factors of 64a3 - 343b 3 are

bearhunter [10]

$\huge\mathfrak\green{Constant}$

Constant is the number that cannot change the value

Im not sure..but you can copy it

8 0

2 years ago

it is claimed that proportion in favor of proportion A is greater than 60%. A sample of size 100 found 69 in favor. what is the

GenaCL600 [577]

Answer:

The alternative hypothesis is $H_1: p > 0.6$ .

The critical value is $Z_c = 1.645$

Step-by-step explanation:

It is claimed that proportion in favor of proportion A is greater than 60%.

This means that at the null hypothesis, we test if the proportion is of at most 60%, that is:

$H_0: p \leq 0.6$

At the alternative hypothesis, we test if the proportion is more than 60%, that is:

$H_1: p > 0.6$

What is (are) the critical value?

The critical value is the value of Z with a p-value 1 subtracted by the standard significance level of 0.05, since we are testing if the mean is more than a value, so, looking at the z-table, $Z_c = 1.645$

7 0

2 years ago