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DanielleElmas [232]
3 years ago
10

State the y-coordinate of the y-intercept for the function below of x^3-9x

Mathematics
2 answers:
uranmaximum [27]3 years ago
6 0

Answer:

The y coordinate of the y-intercept is 0.

Step-by-step explanation:

Consider the provided function.

y=x^3-9x

We need to find y coordinate of y intercept.

y intercepts are the point where the x value is 0 or the line cuts the y axis.

To find the y coordinates of y intercept, substitute x=0 in the provided equation.

y=(0)^3-9(0)

y=0

Therefore, the coordinates are (0,0)

Hence, the y coordinate of the y-intercept is 0

prohojiy [21]3 years ago
4 0

y = x^{3}  - 9x
y = 0^{3}  - 9(0)
y = 0
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Answer:

x = -21

Step-by-step explanation:

62 + 2x - 20 = 0.

Combine like terms

42 +2x =0

Subtract 42 from each side

42-42 +2x = 0-42

2x = -42

Divide each side by 2

2x/2 = -42/2

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Answer: 1/2x -5

Step-by-step explanation:

Half of x , less 5. (5 less just means subtract)

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3 years ago
Which one doesn’t belong? Why? Explain. please help me
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A sample size 25 is picked up at random from a population which is normally
Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

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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.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

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

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

6 0
2 years ago
Answer PLEASE (NO LINKS)
uranmaximum [27]

Answer:

The second answer. If I got it wrong tried my best

Step-by-step explanation:

brainliest

4 0
2 years ago
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