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saw5 [17]
3 years ago
15

Which are solutions to the equation below?

Mathematics
2 answers:
pishuonlain [190]3 years ago
5 0
X^2 + 18 = -9x
x^2 +9x +18 = 0
(x+3) (x+6) =0

So x = -3 and x =-6
Answers: E and F
mojhsa [17]3 years ago
3 0
The answers would be E and F -6 and -3
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Two or zero expresses the possible number of positive real  solutions for the given polynomial equation.

Answer: Option D

<u>Step-by-step explanation:</u>

Given equation:

        x^{3}-4 x^{2}-7 x+28=0

First, we put hit and trial method to find out the one solution. So, if we put x=4 then the above expression will become zero.  We can also write the above expression as  

      (x-4) \times\left(x^{2}-7\right)=0

We know the formula, a^{2}-b^{2}=(a+b)(a-b), make use of this, we get

   (x-4) \times(x-\sqrt{7}) \times(x+\sqrt{7})=0

So, x=4, \sqrt{7},-\sqrt{7}

Hence, from the above expression, we have three values of x as x= 4, 2.64 and -2.64

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The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum
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Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

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 = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

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

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

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

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

X = 25

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

Z = \frac{25 - 26}{0.762}

Z = -1.31

Z = -1.31 has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

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

-1.645 = \frac{X - 26}{0.762}

X - 26 = -1.645*0.762

X = 24.75

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

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

1.645 = \frac{X - 26}{0.762}

X - 26 = 1.645*0.762

X = 27.25

Between $24.75 and $27.25.

3 0
3 years ago
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