While on vacation, your family traveled 2040 miles in 6 days. Your average speed was 340 miles
1 answer:
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See the attached figure to better understand the problem
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
Answer:
a) 68.26% probability that a student scores between 350 and 550
b) A score of 638(or higher).
c) The 60th percentile of test scores is 475.3.
d) The middle 30% of the test scores is between 411.5 and 488.5.
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:
a. What is the probability that a student scores between 350 and 550?
This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So
X = 550
has a pvalue of 0.8413
X = 350
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a student scores between 350 and 550
b. If the upper 3% scholarship, what score must a student receive to get a scholarship?
100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88
A score of 638(or higher).
c. Find the 60th percentile of the test scores.
X when Z has a pvalue of 0.60. So it is X when Z = 0.253
The 60th percentile of test scores is 475.3.
d. Find the middle 30% of the test scores.
50 - (30/2) = 35th percentile
50 + (30/2) = 65th percentile.
35th percentile:
X when Z has a pvalue of 0.35. So X when Z = -0.385.
65th percentile:
X when Z has a pvalue of 0.35. So X when Z = 0.385.
The middle 30% of the test scores is between 411.5 and 488.5.