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Mandarinka [93]
3 years ago
10

(x - 3)(x2 + 3x + 9)

Mathematics
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

x^{3} -5x^{2} +15x -18

Step-by-step explanation:

brainliest

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Which of the following is a Riemann Sum used to estimate?
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A. Area under a curve

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A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean(\mu) = 500 and a standard deviation(\sigma) =  100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = \dfrac{Y-\mu}{\sigma}

Z-score = \dfrac{625-500}{100}

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Therefore , the required z-score = 1.25

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3 years ago
The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.
Artemon [7]

Answer:

P(X \geq 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N(\mu=73,\sigma^{2}=3^{2})

The z score probability distribution is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 73

           \sigma = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X \geq 74)

 P(X \geq 74) = P( \frac{X-\mu}{\sigma} \geq \frac{74-73}{3} ) = P(Z \geq 0.33) = 1 - P(Z < 0.33)

                                               =  1 - 0.62930 = 0.3707                  

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

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3 years ago
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Alisiya [41]

Answer:

i don't know exactly.

Step-by-step explanation:

but the might be an angle tool online.

5 0
2 years ago
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