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3 years ago

2x−9y=14 x=−6y+7 x= y=

Mathematics

1 answer:

Assoli18 [71]3 years ago

7 0

Answer:

x =7, y = 0

Step-by-step explanation:

$2x - 9y = 14...(1) \\ \\ x = - 6y + 7...(2) \\ \\ from \: equation \: (1) \: and \: (2) \\ \\ 2( - 6y + 7) - 9y = 14 \\ \\ - 12y + 14 - 9y = 14 \\ \\ - 21y = 14 - 14 \\ \\ - 21y = 0 \\ \\ y = \frac{0}{ - 21} \\ \\ \huge \red{ y = 0} \\ \\ x = - 6 \times 0 + 7 \\ \\ x = 0 + 7 \\ \\ \huge \purple{x = 7}$

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Which of the following in an identity?

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By "which is an identity" they just mean "which trigonometric equation is true?"

What you have to do is take one of these and sort it out to an identity you know is true, or...

*FYI: You can always test identites like this:
Use the short angle of a 3-4-5 triangle, which would have these trig ratios:
sinx = 3/5 cscx = 5/3
cosx = 4/5 secx = 5/4
tanx = 4/3 cotx = 3/4
Then just plug them in and see if it works. If it doesn't, it can't be an identity!

Let's start with c, just because it seems obvious.
The Pythagorean identity states that sin²x + cos²x = 1, so this same statement with a minus is obviously not true.

Next would be d. csc²x + cot²x = 1 is not true because of a similar Pythagorean identity 1 + cot²x = csc²x. (if you need help remembering these identites, do yourslef a favor and search up the Magic Hexagon.)

Next is b. Here we have (cscx + cotx)² = 1. Let's take the square root of each side...cscx + cotx = 1. Now you should be able to see why this can't work as a Pythagorean Identity. There's always that test we can do for verification...5/3 + 3/4 ≠ 1, nor is (5/3 + 3/4)².

By process of elimination, a must be true. You can test w/ our example ratios:
sin²xsec²x+1 = tan²xcsc²x
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(9/25)(25/16)+1 = (16/25)(25/9)
(225/400)+1 = (400/225)
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(256/144) = (256/144)

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What is the equation of the following line written in general form?

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3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$

$X - 3.6 = 0.4*1.75$

$X = 4.3$

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2x−9y=14 x=−6y+7 ​ x= y=

2x−9y=14 x=−6y+7 x= y=