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egoroff_w [7]
2 years ago
11

Find the slope of the line passing through the points (-7, -4) and (8,8).

Mathematics
2 answers:
nydimaria [60]2 years ago
8 0

Answer:

0.8

Step-by-step explanation:

Slope=(y2-y1) (8--4) (12)

(x2-x1)= (8--7)= (15)= 0.8

lozanna [386]2 years ago
7 0

Answer:

<u>Hope its work for you</u>

Step-by-step explanation:

As we know that,

<h2>slope=m=(Y2-Y1)/(X2-X1) </h2>

As given values

take,   (X1,Y1)=(-7,-4)

           (X2-Y2)=(8,8)

hence   X1=-7  ,  Y1= -4   ,  X2=8  , Y2= 8

putt the values in the above equation of slope we get ,

m=(8-(-4))/(8-(-7)

<h2><u>m=12/15    Ans</u></h2>

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Y=74(1.01)^8 is it increasing or decreasing
Dvinal [7]

Answer:

Increasing

Step-by-step explanation:

The computation is given below:

Given that

y = 74 × (1.01)^8

= 74 × 1.082856706

= 80.1314

As it can be seen that the initial amount is 74 but after solving the given equation the value is increased as it shows 80.1314

Therefore it is increasing

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3 years ago
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Answer:

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Step-by-step explanation:

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4 years ago
a box of cereal contains 23.4 ounces. It cost $5.49. What is the cost,to the nearest cent, of the cereal per ounce.
anyanavicka [17]
Do 5.49 / 23.4
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The answer is $0.23
5 0
3 years ago
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mars1129 [50]
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3 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

3 0
3 years ago
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