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3 years ago

Find the slope the line passing thru (8,4) and (9,-2)

Mathematics

1 answer:

Inessa05 [86]3 years ago

6 0

Answer = -1/6
Solutions in the picture attached

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In a scene from a movie, 60 zombies are following a path through the woods, and they end up at one of

Crank

Answer:

waterfall 5 percent

ravine 30

creek 15percent

cabin 20

hill 5 percent

bluff 10

ravin 25 percent

probability is 98 percent

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3 years ago

Based on our Starbucks analysis report, 4 out of every 20 drinks are returned because of a customer complaint and 16/20 are not

Leni [432]

Answer:

Step-by-step explanation:

the expected profit should be 2 dollars a drink

5 0

1 year ago

Read 2 more answers

Which of the following expresses the distance between the medians as a multiple of the greater interquartile range?

Annette [7]

Answer:

In fact, the distance between the means is or 3.2 kg, over 6 times the larger ... Visually, this means the range of typical values does not overlap.

Step-by-step explanation:

In fact, the distance between the means is or 3.2 kg, over 6 times the larger ... Visually, this means the range of typical values does not overlap.

3 0

3 years ago

What system of equation represents this situation?

Mashutka [201]

Answer:

It's choice A.

Step-by-step explanation:

Kevin is 3 years older than Daniel gives:

k = d + 3.

2 years ago kevin is aged k - 2 and daniel d - 2 years.

So:

k - 2 = 4(d - 2).

8 0

4 years ago

Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou

monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

B) 0.6796

C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = $P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )$

= P ( Z > 2.33 ) = 0.0099

= 0.9%

B) percentage of oranges weighing less than 8.7 ounces

P( x < 8.7 ) = $P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )$

= P ( Z < 0.4667 ) = 0.6796

= 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = $P (\frac{6.2-8}{1.5} < \frac{x - u}{ std} < \frac{7-8}{1.5} )$

= P ( -1.2 < Z < -0.66 )

= Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0

4 years ago