Sven is trying to find the maximum amount of time he can spend practicing the five scales of piano music he is supposed to be wo
1 answer:
Answer:
He should spend 3 minutes or less on each scale
Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal
Step-by-step explanation:
Let
t ------> is the number of minutes he spends on each scale
Remember that the phrase "at least" is equal to "greater than or equal"
so
The inequality that represent this scenario is
solve for t
Multiply by -1 both sides
Divide by 5 both sides
Sven is incorrect
He should spend 3 minutes or less on each scale
Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal
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Complete Question:
On a number line, the coordinates of X, Y, Z, and W are −8, −5, 4, and 6, respectively. Find the lengths of the two segments below. Then tell whether they are congruent. and
Answer:
They are not congruent
Step-by-step explanation:
Length of segment XY:
Coordinate of X = -8
Coordinate of Y = -5
= |-8 -(-5)| = |-8 + 5| = 3
Length of ZW:
Coordinate of Z = 4
Coordinate of W = 6
= |4 - 6| = 2
≠
, therefore, they are not congruent.
Answer:
1) 22.66%
2) 20
Step-by-step explanation:
The scores of a test are normally distributed.
Mean of the test scores = u = 22
Standard Deviation = = 4
Part 1) Proportion of students who scored atleast 25 points
Since, the test scores are normally distributed we can use z scores to find this proportion.
We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:
P(X ≥ 25)
We can convert this value to z score and use z table to find the required proportion.
The formula to calculate the z score is:
Using the values, we get:
So,
P(X ≥ 25) is equivalent to P(z ≥ 0.75)
Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266
Since,
P(X ≥ 25) = P(z ≥ 0.75), we can conclude:
The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%
Part 2) 31st percentile of the test scores
31st percentile means 31%(0.31) of the students have scores less than this value.
This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.
Using the z table, the z score for 31st percentile comes out to be:
z = -0.496
Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.
Using the values, we get:
Thus, a test score of 20 represent the 31st percentile of the distribution.
Answer:
a) R1ºR1 = R1
b) R1ºR2 = R1
c) R1ºR3 =
d) R1ºR4 =
e) R1ºR5 = R1
f) R1ºR6 =
g) R2ºR3 =
h) R3ºR3 = R3
Step-by-step explanation:
R1ºR1
(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1
R1ºR2
This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1
R1ºR3
(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.
R1ºR4
This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.
R1ºR5
If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1
R1ºR6
The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.
R2ºR3
This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.
R3ºR3
If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3
I hope you find this answer useful!