The sample std. dev. will be (14 inches) / sqrt(49), or (14 inches) / 7, or 2 inches.

Find the z score for 93.8 inches:
93.8 inches - 91.0 inches 2.8 inches
z = ------------------------------------- = ----------------- = 1.4
2 inches 2 inches

Now find the area under the standard normal curve to the left of z = +1.4.

My calculator returns the following:

normalcdf(-100,1.4) = 0.919. This is the probability that the mean annual precipitation during those 49 years will be less than 93.8 inches.

4 0

3 years ago

Which equation is correct? arccsc(x)=1/cos(x)

viva [34]

Answer:

arccsc(x)=1/cos(x)

Step-by-step explanation:

7 0

3 years ago

In tossing four fair dice, what is the probability of tossing, at most, one 3

11Alexandr11 [23.1K]

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0

3 years ago

Estimate 5,403 divided by 94

VLD [36.1K]

Answer:

Step-by-step explanation:

7 0

3 years ago

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