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vodomira [7]
3 years ago
9

Maree is setting out snacks for friends she is having over. She has 18 crackers and 9 slices of cheese. If she wants each plate

to be identical, what is the greatest number of plates she can prepare? *
1 point
A. 18 plates
B. 14 plates
C. 9 plates
D. 72 plates
Mathematics
2 answers:
Grace [21]3 years ago
7 0

Answer: C. 9

Step-by-step explanation: If she put 1 cracker and 1 slice of cheese on each plate so it could be identical, she would get the greatest amount of plates she could prepare. This is why the answer is 9.

Katarina [22]3 years ago
3 0
The answer is 9 plates (C)
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4 years ago
13. How much money will 8820 Vield in investa
zavuch27 [327]

9514 1404 393

Answer:

  • 10,247.38 from continuous compounding
  • 10,228.50 from semiannual compounding
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Step-by-step explanation:

The formula for the account balance from continuously compounded interest at annual rate r for t years is ...

  A = Pe^(rt) . . . . P = principal invested

  A = 8820e^(0.05·3) ≈ 10,247.38 . . . continuous compounding

__

The formula for the account balance from interest compounded semiannually at annual rate r for t years is ...

  A = P(1 +r/2)^(2t)

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7 0
3 years ago
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3 0
3 years ago
In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence
Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

\hat p =\frac{12}{300}=0.04

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

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

The critical value of <em>z</em> for 95% confidence level is,

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0
3 years ago
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