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marusya05 [52]
3 years ago
12

The binomial (y − 2) is a factor of y2 − 10y + 16. What is the other factor?

Mathematics
1 answer:
gulaghasi [49]3 years ago
8 0

Answer:

(y - 8)

Step-by-step explanation:

If you factor the original expression, y^2 - 10y + 16, you get (y-2)(y-8) so (y-8) is the other factor. You could also test this by multiplying (y-2)(y-8) together and seeing if you get the original expression.

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The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean
Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 591, \sigma = 42

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{42}

Z = 0.21

Z = 0.21 has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{50}} = 5.94

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{5.94}

Z = 1.515

Z = 1.515 has a pvalue of 0.9351

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{100}} = 4.2

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

Z = \frac{600 - 591}{4.2}

Z = 2.14

Z = 2.14 has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

8 0
3 years ago
Please help hard math lots of points
avanturin [10]

Answer:

C- y=2|x+2|

Step-by-step explanation:

When the number is inside the absolute value bars, the value shifts- so a positive two would move left two, and a negative two would move right two. If the number was outside of the equation, the shape would move up or down- a positive two would move the shape up two, and a negative two would move the graph down two.

8 0
3 years ago
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Question 2 help pls
KengaRu [80]
X=-2
5-4=1
so whatever is square rooted must be equal to 1
1 squared is 1
-2+3 is one
7 0
3 years ago
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Four frogs jumped a total of 185 centimeters how far did each frog jump
Svet_ta [14]
The answer is 46.25cm
5 0
3 years ago
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Expand in logarithmic expression
olchik [2.2K]
ln( \frac{m^{5}}{n^{2}})^{3} =3ln( \frac{m^{5}}{n^{2}})= 3(ln({m^{5})-ln({n^{2}))=3(5ln(m)-2ln(n) = 15*ln(m)-6*ln(n)



The answer is 15*ln(m)-6*ln(n)  
3 0
3 years ago
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