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1 year ago

Find the values of x and y. State which theorem(s) you used.

Mathematics

1 answer:

Usimov [2.4K]1 year ago

5 0

The required simplified value of x and y is 32 and 84 respectively.

Given that,
To find the values of x and y. State which theorem(s) you used.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Vertical opposite angles

3x = 5x - 64
2x = 64
x = 32

Now,

5x - 64 = 5*32 - 64
= 160 - 64

= 96°
Now complementary angle y
y = 180 - 96°

y = 84°

Thus, the required simplified value of x and y is 32 and 84 respectively.

Learn more about simplification here:
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Step-by-step explanation:

d = mV

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DENSITY

Density is defined as mass per unit volume.

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Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

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MASS

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VOLUME

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4 years ago

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Step-by-step explanation:

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3 years ago

A researcher is concerned about the impact of students working while they are enrolled in classes, and she likes to know if stud

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Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

$\bar x = 7.10$ -- sample mean

$\sigma=5$ --- sample standard deviation

$n = 150$ --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

$CI = 90\%$ --- the confidence interval

Calculate the $\alpha$ level

$\alpha = 1 - CI$

$\alpha = 1 - 90\%$

$\alpha = 1 - 0.90$

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Divide by 2

$\frac{\alpha}{2} = 0.10/2$

$\frac{\alpha}{2} = 0.05$

Subtract from 1

$1 - \frac{\alpha}{2} = 1 - 0.05$

$1 - \frac{\alpha}{2} = 0.95$

From the z table. the critical value for $1 - \frac{\alpha}{2} = 0.95$ is:

$z = 1.960$

Solving (c): Margin of error

This is calculated as:

$E = z * \frac{\sigma}{\sqrt n}$

$E = 1.960 * \frac{5}{\sqrt {150}}$

$E = 1.960 * \frac{5}{12.25}$

$E = \frac{1.960 *5}{12.25}$

$E = \frac{9.80}{12.25}$

$E = 0.800$

Solving (d): The confidence interval

This is calculated as:

$CI = (\bar x - E, \bar x + E)$

$CI = (7.10 - 0.800, 7.10 + 0.800)$

$CI = (6.3, 7.9)$

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

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