Solve the given equation

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

5 0

The answer is R=0.053707443....
Round to the nearest decimal place as it asks for
i hope this helps :)

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Write in slope intercept form given two points (1,2) and (-2,5)

Anvisha [2.4K]

I’m pretty sure it’s -1
How you solve this is y2-y1/x2-x1
So 5-2/-3-1 which would be -1

4 0

3 years ago

Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standar

Natali5045456 [20]

Answer:

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 175, \sigma = 22$

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

90th percentile

The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So

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

$1.28 = \frac{X - 175}{22}$

$X - 175 = 22*1.28$

$X = 203.16$

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds

6 0

3 years ago

An 8-sided fair die is rolled twice and the product of the two numbers obtained when the die is rolled two times is calculated.

SCORPION-xisa [38]

Answer:

(a)See below

(b)I) 0.125 (ii)0.828125 (iii)0.234375 (iv) 0.625 (v)0.65625

Step-by-step explanation:

When an 8-sided die is rolled twice, the sample space is the set of all possible pairs (a,b) where a is the 1st outcome and b is the 2nd outcome.

The sample space is:

$[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(1, 7),(1, 8)\\(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(2, 7),(2, 8)\\(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(3, 7),(3, 8)\\(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(4, 7),(4, 8)\\(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5, 6),(5, 7),(5, 8)\\(6, 1), (6, 2), (6, 3), (6, 4)(6, 5),(6, 6),(6, 7),(6, 8)\\(7, 1), (7, 2), (7, 3), (7, 4)(7, 5),(7, 6),(7, 7),(7, 8)\\(8, 1), (8, 2), (8, 3), (8, 4)(8, 5),(8, 6),(8, 7),(8, 8)]$

The sample space of the product a*b of each pair forms the required possibility diagram.

This is given as:

$1, 2, 3, 4, 5, 6,7,8\\2, 4, 6, 8, 10, 12,14,16\\3,6,9,12,15,18,21,24\\4,8,12,16,20,24,28,32\\5,10,15,20,25,30,35,40\\6,12,18,24,30,36,42,48\\7,14,21,28,35,42,49,56\\8,16,24,32,40,48,56,64$