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1 year ago

Can you help me here pls? Asap

Mathematics

1 answer:

boyakko [2]1 year ago

3 0

Step-by-step explanation:

% calculations are totally easy, if you remember to always find and the use 1%.

100% = $2.85

1% = 100%/100 = 2.85/100 = 0.0285

the price difference was

2.91 - 2.85 = $0.06

how many % are these $0.06 compared to yesterday's price ?

that is how many times 1% can fit into this number.

0.06 / 0.0285 = 2.105263158...%

so, rounded this is 2.1%

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"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

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.

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

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

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

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

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

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

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

n= 147.28 ≅ 148 boxes.

I hope it helps!

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Answer:

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Step-by-step explanation:

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