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Brrunno [24]
3 years ago
6

Find the slope and y-intercept for the two linear functions. Linear Functions Identify the slope of the line given in the table:

Identify the y-intercept of the line given in the table: Identify the slope of the line given by the equation: Identify the y-intercept of the line given by the equation:

Mathematics
2 answers:
Natalija [7]3 years ago
5 0

ANSWER

See explanation

EXPLANATION

The table contains these two points (-4,-3) and (2,-4.5)

The slope of this line is given by;

m =  \frac{rise}{run}

m =  \frac{ - 4.5 -  - 3}{2 -  - 4}  =  \frac{ -  1.5}{6}

m =  -  \frac{1}{4}

The y-intercept can be obtained by finding the equation of the line in the form:

y = mx + c

We plug in the slope and the point (-4,-3) to find c.

This implies that,

- 3 =  -  \frac{1}{4}  \times  - 4 + c

- 3 = 1 + c

c =  - 3 - 1 =  - 4

Hence the y-intercept is -4.

The given line is

y =  \frac{3}{2} x + 1

This is already in the slope intercept form:

The slope is

m =  \frac{3}{2}

The y-intercept is c=1

I am Lyosha [343]3 years ago
3 0

Answer:

1. -1/4

2. -4

3. 3/2

4. 1

Step-by-step explanation:

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Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are  (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes n_{1} = 400 and n_{2} = 300

Proportion of mean p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52

<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

<u>Alternative hypothesis H1:</u>- p1 ≠ p2

a) The test statistic is

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }

where p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}  

on calculation we get   p = 0.56    

now q =1-p = 1-0.56=0.44

Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )}  } }\\   =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300}   }

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

<u>conclusion</u>:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) <u>95% confidence intervals</u>

The confidence intervals are P± 1.96(√PQ/n)

we know that = p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals  (0.523 ,0.596)</u>

<u></u>

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