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umka2103 [35]
3 years ago
12

Nancy bought an extra-large bag of 1,288 beads to make necklaces. When she got home, she found another bag with 784 beads in it.

If she uses all of the beads and uses 56 beads per necklace, how many necklaces can she make?
Mathematics
2 answers:
TEA [102]3 years ago
6 0
She can make 37necklaces 
tekilochka [14]3 years ago
5 0

We are given

Nancy bought an extra-large bag of 1,288 beads to make necklaces

When she got home, she found another bag with 784 beads in it

so, total number of beads =1288+784

total number of beads =2072

now, we have

For one necklace , 56 beads are required

so, 56 beads = 1 necklace

1 bead=\frac{1}{56}necklace

for 2072 beads

2072*1 beads=2072*\frac{1}{56}necklace

2072 beads=37necklace

so,

number of necklaces she can make is 37..........Answer

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P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

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P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

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For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

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For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

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Compute the value of P (X ≥ 20) as follows:

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Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

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