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3 years ago

What is the inverse of the following statement? If ∆ABC is equilateral, then it is isosceles.

Mathematics

2 answers:

Nastasia [14]3 years ago

6 0

Answer:"C"

Step-by-step explanation:

Aleksandr-060686 [28]3 years ago

5 0

We have to find the inverse of the given statement:

" If ∆ABC is equilateral, then it is isosceles."

A triangle is said to be equilateral if all of the three sides are equal in measure and a triangle is said to be isosceles if two sides of a triangle are equal in measure.

1. Consider the first option

" If ∆ABC is equilateral, then it is isosceles"

This statement is true but is not the inverse statement of the given statement.

2. Consider the second option

"If ∆ABC is not isosceles, then it is not equilateral."

If triangle ABC is not isosceles, it means the two sides of a triangle does not have an equal measure, which is quiet obvious that it will not be equilateral triangle. As to be an equilateral triangle, all the three sides should be of equal measure.

Therefore, it is the correct inverse statement of the given statement.

3. Consider the third option

"If ∆ABC is not equilateral, then it is not isosceles"

This is not necessarily true, if a triangle is not equilateral, it means that the three sides of a triangle does not have an equal measure. But still two sides of a triangle can still have equal measure. Therefore, it can be isosceles.

Therefore, it is not the correct inverse statement for the given statement.

4. Consider the fourth statement

"If ∆ABC is isosceles, then it is equilateral"

If a triangle is isosceles, it means that the two sides of a triangle have an equal measure. So, it can not be equilateral triangle.

Therefore, it is not the correct inverse statement for the given statement.

So, Option 2 is the correct answer.

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4 years ago

A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

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

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

<em>Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.</em>

So, 95% confidence interval for the population proportion, p 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

<u>95% confidence interval for p</u> = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]

= [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

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ikadub [295]

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Step-by-step explanation:

Plug in (x+3) in the F(x) function so:

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