Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
C) Between 80 and 120
D) less than 80
E) Between 70 and 100
F) More than 130
If they're travelling towards each other;
Effective speed = 75+55 = 130
Distance = 338m
Time= distance/ speed
T= 338/130
T= 2.6 hrs
The answer would be: t= -5/4
Answer:
3.14
Step-by-step explanation:
Given that:
Mean weight (m) = 79 ounces
Population standard deviation (s) = 13.1
Sample size = 47
Maximal margin of error associated with 90% confidence interval.
The margin of error is given by:
Zcritical * (standard deviation / sqrt(sample size)
Z critical at 90% confidence interval = 1.645
Hence,
Zcritical * (standard deviation / sqrt(sample size)
1.645 * 13.1 / sqrt(47)
1.645 * (13.1 / 6.8556546)
1.645 * 1.9108313
Hence, the margin of error is :
3.1433174885
= 3.14
Answer:
? = -1 1/4 (Negetive one and one fourth)
Step-by-step explanation:
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