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3 years ago

What is the slope of this line?

Mathematics

2 answers:

OLga [1]3 years ago

6 0

The slope of this line is -2.

rise/run

rise = 4
run = -2

4
------- = -2
-2

Using the points (1,0) & (0,2)

2 - 0 2
----------- = ----- = -2
0 - 1 -1

irina [24]3 years ago

6 0

Since this line is descending, its slope should be negative:
Moreover the slope is nothing than the tangent of the angle formed with x-axis.
& the tangent of Ф = adjacent side/ opposite side).
If read well the graph you see that tan Ф = (4 units)/( - 8 units) = -4/8 = -1/2
Hence the slope = tan = -1/2

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6) Find the difference of (4x - 5) and (6x + 8)

Allushta [10]

Answer:

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Step-by-step explanation:

Hi there!

$(4x - 5)-(6x + 8)$

Open up the parentheses

$4x - 5-6x - 8$

Combine like terms

$4x -6x- 5 - 8\\-2x- 13$

I hope this helps!

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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> = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

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