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N76 [4]
3 years ago
10

Pls help and thank youuu

Mathematics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

slope= 3/2

Step-by-step explanation:

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2. Select the situation below that result in zero. * 0 points A. A diver is at sea level, reaches a depth of 6 ½ feet and descen
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Answer: B and C only

Step-by-step explanation:

The temperature on a cold day starts at 0 degrees Celsius, falls 4 degrees, and rises 4 degrees

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(Initial temperature + rise in temperature + fall in temperature)

Summing all : (0 + 4 - 4) = 0

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3 years ago
Consider a uniform distribution from aequals4 to bequals29. ​(a) Find the probability that x lies between 7 and 27. ​(b) Find th
weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

P(c \leq x \leq d) = \frac{d - c}{b - a}

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So c = 7, d = 27

P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8

80% probability that x lies between 7 and 27.

​(b) Find the probability that x lies between 6 and 13. ​

So c = 6, d = 13

P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

​So c = 9, d = 20

P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So c = 11, d = 18

P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28

28% probability that x lies between 11 and 18.

3 0
3 years ago
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ahrayia [7]

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