Answer:
p=-54
Step-by-step explanation:
Whats the quotient of 8453 and 35?
1 answer:
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Answer:
<h3> To find x you need to divide polygon into triangles</h3><h3>The formula : "the sum of exterior angles=360°" is false for concaved polygons</h3>Step-by-step explanation:
α = 180° - 135° = 45°
β = 180° - 120° = 60°
γ = 180° - 60° = 120°
δ = 180° - x
ε = 180° - 80° = 100°
If we divide the pentagon into five triangles then:
5•180° = α + β + γ + δ + ε + 4x
900° = 45° + 60° + 120° + x + 100° + 4x
900° = 325° + 5x
5x = 575°
x = 115°
60° + 120° + 135° + 80° + 115° = 510° ≠ 360°
Answer:
a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean
Step-by-step explanation:
Problems of normally distributed distributions 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 question, we have that:
Probabilities:
A) Between 55,000 and 65,000 miles
This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So
X = 65000
has a pvalue of 0.954
X = 55000
has a pvalue of 0.677
0.954 - 0.677 = 0.277
0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
B) Less than 48,000 miles
This is the pvalue of Z when X = 48000. So
has a pvalue of 0.348
0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
C) At least 41,000 miles
This is 1 subtracted by the pvalue of Z when X = 41,000. So
has a pvalue of 0.108
1 - 0.108 = 0.892
0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
D) A lifetime that is within 10,000 miles of the mean
This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So
X = 61200
has a pvalue of 0.889
X = 41200
has a pvalue of 0.111
0.889 - 0.111 = 0.778
0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean