Considering only the values of β for which (1+cosβ)(1−cosβ)sinβ is defined, which of the following expressions is equivalent to
1 answer:
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Answer:
a)
b)
c) 40 weeks gestation babies
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.
Here, we use . So
b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.
Here, we use . So
c. For which group is a birth length of 45 cm more common?
For each group, the probability is 1 subtracted by the pvalue of Z.
Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.
= 1455
generate a few terms of the sequence using = 3n + 2
= ( 3 × 1) + 2 = 5
= (3 × 2) + 2 = 8
= (3 × 3 ) + 2 = 11
= (3 × 4 ) + 2 = 14
= ( 3 × 5 ) + 2 = 17
the terms are 5, 8, 11, 14, 17
these are the terms of an arithmetic sequence
sum to n terms is calculated using =
[ 2a + (n-1)d]
where a is the first term and d the common difference
d = 8 - 5 = 11 - 8 = 14 - 11 = 3 and = 5
=
[( 2 × 5) + (29 × 3) ]
= 15( 10 + 87) = 15 × 97 = 1455