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3 years ago

What is the converse of the statement?Perpendicular lines have negative reciprocal slopes.

Mathematics

2 answers:

Alexxandr [17]3 years ago

6 0

Answer:

The converse of the statement is:

If the Lines have negative slope than the lines are perpendicular.

Step-by-step explanation:

" In logic, the converse of a categorical or implicational statement is the result of reversing its two parts. For the implication P → Q, the converse is Q → P "

We are given implicational statement as:

Perpendicular lines have negative reciprocal slopes.

i.e. we can write this statement as:

If lines are Perpendicular than lines have negative slope.

Here we have:

P= Line are perpendicular

Q= lines have negative slope

Now the converse of the statement is:

If the Lines have negative slope than the lines are perpendicular

77julia77 [94]3 years ago

4 0

The converse is when u switch the hypothesis and the conclusion.

so the converse of this statement would be :
if they have negative reciprocal slopes, then they are perpendicular lines.

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HELP ASAP!! 100 POINTS!! describe how you would plot the point (-4, 3)

fredd [130]

Answer:

Hi!

First you would start at -4 on the x-axis, then you would move up 3 spaces and put your point there.

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I hope this helps you!

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3 years ago

Read 2 more answers

Solve (2 - x)^2 < 4/25

Semenov [28]

Answer:

8/5 < x< 12/5

Step-by-step explanation:

(2 - x)^2 < 4/25

Take the square root of each side

sqrt((2 - x)^2) <±sqrt( 4/25)

Make two equations

2-x < 2/5 2-x > -2/5

Subtract 2 from each side

2-x-2 < 2/5 -2 2-x-2 > -2/5-2

-x < 2/5 - 10/5 -x > -2/5 - 10/5

-x < -8/5 -x > -12/5

Multiply by -1, remembering to flip the inequality

x> 8/5 x < 12/5

8/5 < x< 12/5

5 0

3 years ago

Read 2 more answers

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

You invest $5,000 in a savings account that earns 6% interest each year.

weqwewe [10]

A. Multiply $5000 by .06, and add it back to $5000 to get the total after one year. This is $5300.
B. Multiply the new amount, or $5300 by .06, and add it back to $5300 to get the total after 2 years. This is $5618.

A. $5300
B. $5618

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3 years ago

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Vera_Pavlovna [14]

There’s an app called photo math where u can take a picture of the problem and it gives u the answer.

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3 years ago