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1 year ago

Which are correct representations of the inequality -3(2x - 5) <5(2 - x)? Select two options.

Mathematics

2 answers:

Alex17521 [72]1 year ago

6 0

Answer: The answer is C and E

Step-by-step explanation:

Hope this helps;)

Sonja [21]1 year ago

5 0

Answer:

The third and fourth options.

Step-by-step explanation:

You simplify the inequality first...

-6x + 15 < 10 - 5x

-x < -5

x > 5

The first option is incorrect since it is less than.

The second option is basically -5x < 15; x > -3; that's incorrect.

The third option is basically -x < -5; x > 5; that is correct!

The fourth option is correct since it shows more than 5.

The fifth option is incorrect because it shows less than -5.

Hope this helps!

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astraxan [27]

Answer:

She should buy the monthly plan for the unlimited movies rather than pay $2.99 per movie. This is because, the more she pay that amount for each movie, the higher her expenses would become at the end of each month.

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That notwithstanding other days which will feel like watching movies or the public holidays which she would be free to relax.

Step-by-step explanation:

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2 years ago

1. & 2. In the diagram below, points A, B, and C are collinear. Answer each of the following questions. The figure shown bel

KiRa [710]

Answer:

a). AB = 8 in

b). AB = 9.75 in

c). AC = 6.5 in

d). BC = 1.5 in

Step-by-step explanation:

a). Since, AB = AC + CB

Length of AC = 5 in. and CB = 3 in.

Therefore, AB = 5 + 3 = 8 in.

b). Given : AC = 6.25 in and CB = 3.5 in

Therefore, AB = AC + CB = 6.25 + 3.5

AB = 9.75 in.

c). Given: AB = 10.2 in. and BC = 3.7 in.

AB = AC + BC

AC = AB - BC

AC = 10.2 - 3.7

AC = 6.5 in

d). Given: AB = 4.75 in and AC = 3.25 in.

BC = AB - AC

BC = 4.75 - 3.25 = 1.5 in.

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2 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

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2 years ago

A translator earns $9.50 for each page that she translates. If she works 6.5 hours and translates 16 pages each day on average,

Ludmilka [50]

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1 year ago

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2 years ago