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anyanavicka [17]
3 years ago
13

Fred buys flags from a manufacturer for (unknown) dollars each then sells the flags in his store for a 26% markup.

Mathematics
2 answers:
malfutka [58]3 years ago
8 0
The markup is 1.26 (the amount added is .26)

1.26x (when x is the price Fred buys them for from the manufacturer)

1.26(40)
$50.40 is the retail price


50.4-40 
$10.40 is the markup
WINSTONCH [101]3 years ago
6 0

Answer:

The 26% markup means that he is selling the flags 26% above the cost.

So, the cost would represent 100%, which is 1, and 26% of markup in decimal number is 0.26.

The price including the markup would be 1.26 as decimal or 126% as percentage.

The problem states that the cost is unknown, so we represent that with a variable <em>x. </em>The retail price would be 1.26x, which is 126% per flag.

Finally, if Fred paid $40 per flag, the actual retail price would be:

1.26($40)=$50.40.

With $10.40 of markup.

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What is the value of the 11th term in the following geometric sequence? -5, -10, -20, -40
yaroslaw [1]

You are multiplying 2 to each term each time.

-40 is the 4th term

-40 x 2 = -80

-80 x 2 = - 160

-160 x 2 = -320

-320 x 2 = -640

-640 x 2 = -1280

-1280 x 2 = -2560

-2560 x 2 = -5120

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-5120 is your 11th term.

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hope this helps

3 0
3 years ago
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It costs $5.08 for 4 pounds of gummy candy in bulk. How much would it cost to buy 17 pounds?
Daniel [21]

Answer:

$21.59

Step-by-step explanation:

Basically how I did it was to multiply 5.08 by 4, because that would be the cost of 16 bags and then I divided 5.08 by 4 to see how much one bag would cost. One bag costs 1.27. After I've got the answer, I added 1.27 to 20.32 and got $21.59.

5 0
3 years ago
Which expression should you simplify to find the 90% confidence interval,
bogdanovich [222]

Answer:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.2870.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

Step-by-step explanation:

The estimated proportion of interest is \hat p=0.45

We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025.

And the critical value is:z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the true population proportion is interest is given by this formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing the values provided we got:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287\\\\0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

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3 years ago
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s344n2d4d5 [400]

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2 years ago
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Please don’t answer just for points! Help is really needed~
Galina-37 [17]

i dont understand

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