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Paladinen [302]

3 years ago

5

The original price of a bicycle was $100.00. In March, a bicycle went on sale for 10% off. In April, the sale price was raised b

y 10%. What is the price of the bicycle in April? Write just the value that represents your answer.

1 answer:

trapecia [35]3 years ago

5 0

Answer:

take 10% out of 100 = 90

then find 10% of 90 = 9

add that to 90 = 99

The price of the bike in april is 99$

psl mark me as brailyest:)

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Vinil7 [7]

Answer:

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We want to calculate the bounds of a 95% confidence interval of the difference between proportions.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

$p_1=X_1/n_1=494/1000=0.494$

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

$p_2=X_2/n_2=552/1000=0.552$

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$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{494+552}{1000+1000}=\dfrac{1046}{2000}=0.523$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.523*0.477}{1000}+\dfrac{0.523*0.477}{1000}}\\\\\\s_{p1-p2}=\sqrt{0.000249+0.000249}=\sqrt{0.000499}=0.022$

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.022=0.0438$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.058-0.0438=-0.102\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.058+0.0438=-0.014$

The 95% confidence interval for the difference between proportions is (-0.102, -0.014).

7 0

3 years ago