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ivolga24 [154]
4 years ago
9

7x + 2y = -1 3x – 4y = 19

Mathematics
2 answers:
jeka944 years ago
7 0

Answer:  Y = - 4 & x = 1

Step-by-step explanation:

Systems of linear equations: We will solve this using substitution but you can use 2 more ways.

7x + 2y = -1

2x - 4y = 19

Leave X  alone first: -------------------

10x - 2y = 18

10x/10 = 18 + 2y/10

x = 9/5 + y/5

Solve for y:

7(9/5 + y/5) + 2y = -1

63/5 + 7y/5 + 2y - 1

63/5 + 7y/5 = 10y/5 = -1

63/5 + 17y/5 = -1 (-5/5)

17y/5 = -68/5 (cancel the 5)

y = -68/17

y = - 4 (Your Answer)

7x + 2(- 4) = -1

7x + (-8) = -1

7x = 7

x =   1

CHECK=

7(1) + 2(-4) = -1

7+ (-8) = -1.

-1=-1

navik [9.2K]4 years ago
3 0

Answer:

(1, -4).

(x = 1, y = -4).

Step-by-step explanation:

7x + 2y = -1

3x  – 4y = 19    Multiply the first equation by 2:

14x + 4y = -2     Adding:

17x = 17

x = 1.

Plug this into the second equation:

3(1) - 4y = 19

-4y =  19 - 3 = 16

y = -4.

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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.

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Where is the question?

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