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3 years ago

PLEASE HELP! Will mark brainliest! Find the slope of the line:

Mathematics

2 answers:

balandron [24]3 years ago

8 0

Answer:

the final answer is 4/5

Step-by-step explanation:

well the slope formula is the following:

m= (change in y)/(change in x)

let's take the two following points

(-3,0) and (2,4) so

x_1=-3\\x_2=2\\y_1=0\\y_2=4

change in y = y_2-y_1=4-0

change in x = x_2-x_1=2-(-3)=2+3=5

m=4/5

adelina 88 [10]3 years ago

4 0

Answer:

the slope is 4/5

Step-by-step explanation:

well the slope formula is the following:

m= (change in y)/(change in x)

let's take the two following points

(-3,0) and (2,4) so

$x_1=-3\\x_2=2\\y_1=0\\y_2=4$

change in y = $y_2-y_1=4-0$

change in x = $x_2-x_1=2-(-3)=2+3=5$

m=4/5

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Step-by-step explanation:

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3 years ago

Arun is twice as old as Shreet 5 years ago his age was three times shree age find their present ages

Mariulka [41]

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Step-by-step explanation:

In the system of equations Arun's age is a. Shree is s

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2 years ago

Read 2 more answers

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

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