Answer:
His jump was of 272.45 inches
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
75th percentile
X when Z has a pvalue of 0.75. So X when Z = 0.675
His jump was of 272.45 inches
Answer:
Step-by-step explanation:
Let x represent the number of lawns mowed and y represent the number of cars washed.
1. Since the number of cars that Beth wash is no more than four times the number of lawns Mike has scheduled to mow, you have that 
.
2. Beth will wash at least 50 cars, then 
3. Mike charges $25 each time he mows a yard, then he earns $25x for mowing x yards. Beth charges $15 for each car she washes, then she earns $15y for y cars washed. They need at least $1975, so
4. A set of constraints to model the problem is
If we put that into an equation, we get
(5/9)(f/32)
because "five ninths" is a fraction, "times" is multiplication (which can be represented by parenthesis), "the difference" is division, and "the difference of f and 32" is a fraction.
If < 6 = 57, then < 4 = 57.....(corresponding angles are equal)
and < 4 + < 1 = 180.....because they form a linear pair
57 + < 1 = 180
< 1 = 180 - 57
< 1 = 123 <===
Answer:17.3964 males 7.6035 females
Step-by-step explanation:
The first way to calculate the expected uses the marginal percentages. If sex is not related to flavor preference, you would expect the same percentages of males to prefer the same flavors as females. Since overall 20.71% prefer vanilla, you would expect 20.71% of males to prefer vanilla. Now, 20.71% of 84 people is 17.3964, so this is how many males you would expect to prefer vanilla. You would also expect 20.71% of the females to prefer vanilla, which is 17.6035. So you'd expect 17.3964 males and 17.6035 females to prefer vanilla.
