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Angelina_Jolie [31]

2 years ago

12

Find the difference between the actual and estimated values of 55 - 21

2 answers:

Ivahew [28]2 years ago

7 0

Answer:

Actual: 34 Estimated: 40

Step-by-step explanation:

The difference between these values is that the actual difference is more accurate. If you're asking for the difference as in subtracting, then 7 is the answer. Hope this helps :)

Tomtit [17]2 years ago

4 0

Answer:

6

Step-by-step explanation:

Actual values = 55 - 21

= 34

Estimated values = 60 - 20 = 40

Difference between actual and estimated values = 40 - 34 = 6

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13%

Step-by-step explanation:

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3 years ago

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Prove identity trigonometric equation

irina [24]

Explanation:

The given equation is False, so cannot be proven to be true.

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Perhaps you want to prove ...

$2\tan{x}=\dfrac{\cos{x}}{\csc{(x)}-1}+\dfrac{\cos{x}}{\csc{(x)}+1}$

This is one way to show it:

$2\tan{x}=\cos{(x)}\dfrac{(\csc{(x)}+1)+(\csc{(x)}-1)}{(\csc{(x)}-1)(\csc{(x)}+1)}\\\\=\cos{(x)}\dfrac{2\csc{(x)}}{\csc{(x)}^2-1}=2\cos{(x)}\dfrac{\csc{x}}{\cot{(x)}^2}=2\dfrac{\cos{(x)}\sin{(x)}^2}{\cos{(x)}^2\sin{(x)}}\\\\=2\dfrac{\sin{x}}{\cos{x}}\\\\2\tan{x}=2\tan{x}\qquad\text{QED}$

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4 0

3 years ago

Please Help!<br> Write a equation for each hanger!

julsineya [31]

Answer:

Step-by-step explanation:

A: 2x = 3

B: w+1.3 = 2.7

C: 3y = 5.1

D: z+⅓ = ½

7 0

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

8_murik_8 [283]

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(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

$\alpha = 1 - CI$

$\alpha = 1 - 90\%$

$\alpha = 1 - 0.90$

$\alpha = 0.10$

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

6 0

3 years ago

which statement is true about the equation 3/4z-1/4z+3=2/4z+5. It has no solution. It has one solution. It has two solutions. It

kozerog [31]

3/4z - 1/4z + 3 = 2/4z + 5
3/4z - 1/4z - 2/4z = 5 - 3
2/4z - 2/4z = 2
0 ≠ 2
----------------------------
<span>It has no solution.</span>

3 0

3 years ago

Read 2 more answers