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3 years ago

Cuantos terminos tiene -3xy

Mathematics

1 answer:

katrin [286]3 years ago

3 0

Answer:

El a+b tiene 1 terminó porque para separar un término el factor es el +Step-by-step explanation:

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Given tantheta = 4/3 and pi < theta < 3pi/2; find cos2theta

Sidana [21]

Answer:

d. -7/25

Step-by-step explanation:

π<θ<3π/2 in third quadrant

tanθ = 4/3

sinθ = - 4/5

cosθ = - 3/5

cos 2θ = cos²θ - sin²θ = (- 3/5)² - (- 4/5)² = 9/25 - 16/25 = - 7/25

8 0

3 years ago

Read 2 more answers

2x + 11 = 25 Solve x I need the working out.

BlackZzzverrR [31]

First to find what’s 2x you do 25-11=14, and you know that 7 x 2 =14.
So x= 7.

4 0

3 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

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

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

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

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