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goldfiish [28.3K]

2 years ago

8

eric jogged 31/4 miles on monday 55/8 miles on tuesday and 8 miles on wensday .suppose he continues the pattern for the remainde

r of the week .how far will eric jog on friday?

1 answer:

JulijaS [17]2 years ago

5 0

The answer is 55/8 hope you pass:D

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Which is an ordered pair on the line 2.5,14 2.25,12 1.25,5 1.5,9

Savatey [412]

I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25

What is the eqn of the line: y = mx + b becomes

14 = (8)(2.5) + b; find b:

14-20 = b = -6. Then, y = 8x - 6.

Now determine whether (12,1.25) lies on this line.

Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).

Why not work this problem out yourself using my approach as a guide?

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3 years ago

a recent survey of 300 pet owners that have either cats or dogs reveals that two thirds are dog owners. if there are 20,000 regi

alexandr1967 [171]

40,000 Dog owners :3

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3 years ago

Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

$P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175$

Positive when there is raritanium:

98% of 13%

$P(A) = 0.98*0.13 = 0.1274$

What is the probability the rock sample actually contains raritanium?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967$

0.967 = 96.7% probability the rock sample actually contains raritanium

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2 years ago

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valkas [14]

Answer:

B

Step-by-step explanation:

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Answer:

51,040

Step-by-step explanation:

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