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grin007 [14]
2 years ago
14

Y ³ ——— 10y + 4 i got 6/24 or just 4 is this correct

Mathematics
1 answer:
Helga [31]2 years ago
7 0

2 to the power of 3 is

2 x 2 x 2 = 8

10 to the power of 2 is

10 x 10 = 100

then add 4

104

8/104

I believe this is how you do it :) hope this helps

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Assume that the empirical rule is a good model for the time it takes for all passengers to board an airplane. The mean time is 4
slamgirl [31]

Answer:

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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:

\mu = 48, \sigma = 4.

Which interval describes how long it takes for passengers to board the middle 95% of the time?

This is between the 2.5th percentile and the 97.5th percentile.

So this interval is the value of X when Z has a a pvalue of 0.025 and the value of X when Z has a pvalue of 0.975

Lower Limit

Z has a pvalue of 0.025 when Z = -1.96. So

Z = \frac{X - \mu}{\sigma}

-1.96 = \frac{X - 48}{4}

X - 48 = -1.96*4

X = 40.16

Upper Limit

Z has a pvalue of 0.975 when Z = 1.96. So

Z = \frac{X - \mu}{\sigma}

1.96 = \frac{X - 48}{4}

X - 48 = 1.96*4

X = 55.84

The interval that describes how long it takes for passengers to board the middle 95% of the time is between 40.16 minutes and 55.84 minutes.

7 0
3 years ago
What doubles fact can I write to solve 4+5
GenaCL600 [577]
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5 0
3 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
2 years ago
Find the area please :)
Delicious77 [7]

Answer:

it's area is 1/2 b×h

1/2×7×3.8

13.3

4 0
2 years ago
Read 2 more answers
10. A triangle has sides with lengths 3,5, 7. Is this an acute, obtuse, or right<br> triangle?
satela [25.4K]

Answer:

OBTUSE

Step-by-step explanation:

You would first have to figure this put via Pythagorean theroem. Basically do this (attachment)***C IS THE LARGEST NUMBER*** and know that c^2 < a^2 + b^2 (less than) = ACUTE, c^2 > a^2 + b^2 (greater than) = OBTUSE, and c^2 = a^2 + b^2 is RIGHT. If it is Right with all different numbers, it is SCALENE.

8 0
3 years ago
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