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2 years ago

Roberto wants to display his 18 sports cards in an album . Some pages hold 2 cards and others hold 3 cards .

Mathematics

1 answer:

nexus9112 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

Assuming Roberto wants to completely fill each page that he puts cards in, this function describes the number of 2-card pages, a, and 3-card pages, b.

2a + 3b =18

Ricardo can fill up 9 2-card pages, and 6 3-card pages.

a=9, b=0

We must add 2 3-card pages at a time,so that we have an even number for the 2-card pages:

a=6, b=2

Add 2 to b once more:

a=3, b=4

One more time:

a=0, b=6:

Thus, Ricardo can display his figures in the following page combinations:

a=9, b=0

a=6, b=2

a=3, b=4

a=0, b=6

Remember that a= number of 2-card pages and b=number of 3-card pages

There are 4 different ways that Ricardo can arrange his figures in terms of what kind of pages he uses.

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2 years ago

What is 12 1/3-10 11/12

Brut [27]

Answer:

1 5/12

Step-by-step explanation:

12 1/3 - 10 11/12

We need to get a common denominator of 12

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We need to borrow from the 12 (the whole number) because the 2nd fraction is bigger than the first

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2 years ago

12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th

atroni [7]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

$Confidence= 0.95$ the confidence level desired

$ME =0.03$ represent the margin of error desired

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The confidence level is 95% or 0.95, the significance is $\alpha=0.05$ and the critical value for this case using the normal standard distribution would be $z_{\alpha/2}=1.96$

Since we don't have prior information we can use $\hat p= 0.5$ as an unbiased estimator

Also we know that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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2 years ago

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blagie [28]

This is complicated because I’m typing on a phone, but
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2 years ago