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2 years ago

94 POINTS AND BRAINLIEST TO BEST ANSWER!!!

Mathematics

2 answers:

Schach [20]2 years ago

7 0

Answer:

Step-by-step explanation:

Diagonals are perpendicular so

2x+ 90 + 3x = 180 since this forms a triangle

Combine like terms

5x+ 90 = 180

Subtract 90 from each side

5x +90-90 =180-90

5x = 90

Divide each side by 5

5x/5 = 90/5

x =18

We want to find Angle ABD

ABD = 3x = 3*18 = 54

nika2105 [10]2 years ago

4 0

Answer:

HOPE THIS HELPS PLS DON"T TAKE IT DOWNN

Step-by-step explanation:

Say that the intersecting point of both diagonals is O

that means that in triangle AOB, all angles add up to 180º that is just the definition of sum of angles in a triangle

and the m<AOB = 90º, diagonals in a rhombus are perpendicular

that means that

90º + 2x + 3x = 180º

90º + 5x = 180º

5x = 90º

x = 18º

if m<ABD = 3x

then its equal to m<ABD = 3(18)

m<ABD = 54º

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2 years ago

Complete the following statement. 3(4 x 8) (3 x4) (_)

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Answer:

Step-by-step explanation:

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3 years ago

Which fraction is larger 7/16 or 51/100?

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3 years ago

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3 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].

7 0

3 years ago