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Genrish500 [490]
3 years ago
7

Which two values are equivalent to 73 x 7-6?

Mathematics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

Yor answer that is correct is A.

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9x^2-49
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Ricardo earns a base salary of $500 a week plus a weekly commission he has earned a commission of $325, $460, $280, $400, $380,
Vinvika [58]

Answers:

1)

Mean = $349.2

Median = $352.5

Range = $210

2)

$499.90

3)

$355

I have attached a photo of the work written for this question:

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3 years ago
The sum of 14 and 3<br> divided by r
labwork [276]

Answer:

?

Step-by-step explanation:

7 0
3 years ago
the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai
ipn [44]

Answer:

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

Step-by-step explanation:

Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.

So, let x be the variable describing the marks of a student, and

z=(x-\mu)/\sigma

the standardized equivalent of that (mu - mean, sigma - standard deviation).

We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:

z-score for the 10% point: z_10=1.28

z-score for the 20% point: z_20=-0.84

That gives us two equations:

z_{10}=1.28=(75-\mu)/\sigma\\z_{20}=-0.84=(40-\mu)\sigma

and can be solved for mu and sigma (do the work on your end, I am showing my result):

\mu=53.87\,\,,\,\, \sigma=16.51

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

3 0
3 years ago
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