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Rashid [163]
3 years ago
14

Seven pieces of cardstock sell for $3. Tanesha spends $57 on cardstock. How many pieces of cardstock does Tanesha buy?

Mathematics
1 answer:
Helen [10]3 years ago
8 0

Hey there! :)

Answer:

133 pieces.

Step-by-step explanation:

If 7 pieces of cardstock sell for $3, we can set up a proportion to determine the amount of pieces bought in pieces/ money spent:

\frac{7}{3} = \frac{x}{57}

Cross multiply:

7 · 57 = 3 · x

399 = 3x

Divide both sides by 3:

x = 133 pieces. This is the amount of pieces bought.

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3 years ago
A publisher needs to send many books to a local book retailer and will send the books in a combination of small and large boxes.
Paladinen [302]

Answer:

Answer:

\{ {{20x+30y=280} \atop {y=4x}} .{

y=4x

20x+30y=280

.

Where xx is the number of small boxes sent and yy is the number of large boxes sent.

Step-by-step explanation:

Let be xx the number of small boxes sent and yy the number of large boxes sent.

Since each small box can hold 20 books (20x20x ), each large box can hold 30 books (30y30y )and altogether can hold a total of 280 books, we can write the following equation to represent this:

20x+30y=28020x+30y=280

According to the information provided in the exercise, there were 4 times as many large boxes sent as small boxes. This can be represented with this equation:

y=4xy=4x

Therefore, the system of equation that be used to determine the number of small boxes sent and the number of large boxes sent, is:

\{ {{20x+30y=280} \atop {y=4x}} .{

y=4x

20x+30y=280

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7 0
3 years ago
A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From
lutik1710 [3]

Answer:

a) z = 1.645

b) The should sample at least 293 small claims.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645, which means that the answer of question a is z = 1.645.

Now, find  the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

M = 12, \sigma = 124.88. So

M = z*\frac{\sigma}{\sqrt{n}}

12 = 1.645*\frac{124.88}{\sqrt{n}}

12\sqrt{n} = 205.43

\sqrt{n} = \frac{205.43}{12}

\sqrt{n} = 17.12

\sqrt{n}^{2} = (17.12)^{2}

n = 293

The should sample at least 293 small claims.

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