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3 years ago

Sin(x−y)/sinxsiny= coty - cotx

Mathematics

1 answer:

andrew-mc [135]3 years ago

4 0

$x = 0 \\ \\ \sin^{2}(y) \sin( \frac{x - y}{ \sin(x)} ) - \sin(x) \cos(y)\cos(x)\sin(y)= 0$

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How do you describe the solution set to a system of linear inequalities?

Step-by-step explanation:

The solutions of a system of linear inequalities are all the ordered pairs that make all the inequalities in the system true.

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2 years ago

In triangle EFG , EF=FG . EF= 6X-10 , EG =2X+14, and FG =3X+11. find EF .

jarptica [38.1K]

Since EF=FG, you can set 6x - 10 = to 3x + 11

$6x -10 = 3x +11$

Then add like terms

$6x - 3x = 11+10$

so, $3x = 21$

Now you can divide by 3 to get $x = 7$ :)

then put it back into $6x - 10$

$6(7) - 10 = EF$

so, $EF = 32$

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2 years ago

Read 2 more answers

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

Sin(x−y)/sinxsiny= coty - cotx​

Sin(x−y)/sinxsiny= coty - cotx