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3 years ago

Solve this equation.

Mathematics

1 answer:

Anika [276]3 years ago

5 0

Answer:

Answer: b = 58/5 or 11 3/5 or 11.6 decimal

Step-by-step explanation:

Solve for b:

b - (3 + 2/5) = 8 + 1/5

Put 3 + 2/5 over the common denominator 5. 3 + 2/5 = (5×3)/5 + 2/5:

b - (5×3)/5 + 2/5 = 8 + 1/5

5×3 = 15:

b - (15/5 + 2/5) = 8 + 1/5

15/5 + 2/5 = (15 + 2)/5:

b - (15 + 2)/5 = 8 + 1/5

15 + 2 = 17:

b - 17/5 = 8 + 1/5

Put each term in b - 17/5 over the common denominator 5: b - 17/5 = (5 b)/5 - 17/5:

(5 b)/5 - 17/5 = 8 + 1/5

(5 b)/5 - 17/5 = (5 b - 17)/5:

(5 b - 17)/5 = 8 + 1/5

Put 8 + 1/5 over the common denominator 5. 8 + 1/5 = (5×8)/5 + 1/5:

(5 b - 17)/5 = (5×8)/5 + 1/5

5×8 = 40:

(5 b - 17)/5 = 40/5 + 1/5

40/5 + 1/5 = (40 + 1)/5:

(5 b - 17)/5 = (40 + 1)/5

40 + 1 = 41:

(5 b - 17)/5 = 41/5

Multiply both sides of (5 b - 17)/5 = 41/5 by 5:

(5 b - 17)/(5 1/5) = 1/5×1/(1/5) 41

1/5×1/(1/5) = 1:

5 b - 17 = 1/5×1/(1/5) 41

1/5×1/(1/5) = 1:

5 b - 17 = 41

Add 17 to both sides:

5 b + (17 - 17) = 17 + 41

17 - 17 = 0:

5 b = 41 + 17

41 + 17 = 58:

5 b = 58

Divide both sides of 5 b = 58 by 5:

(5 b)/5 = 58/5

5/5 = 1:

Answer: b = 58/5 or 11 3/5 or 11.6 decimal

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Answer:

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Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]

Where

p' is the sample proportion/point estimator of the population proportion

$Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ is the margin of error (d) of the confidence interval.

$Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648$

$d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$

$d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }$

$(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}$

$n*(d*Z_{1-\alpha /2})^2= p'(1-p')$

$n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}$

$n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}$

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I hope it helps!

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Step-by-step explanation:

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