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2 years ago

15

Carla is making bracelets she wants to put 10 charms on each bracelet charms are sold in packets of 12 what is the least number

of packages she can buy to make bracelets with no charms left over

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

Carla can buy 5 packets of charms.

Step-by-step explanation:

5 packets of charms would be able to make 6 bracelets.

*please give me brainliest, i only need three more*

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If g( n ) = 3 n - 2, then what is the value of g(10)?

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12. If n(A) = 14, n(B) = 15, and n(<br>AB) = 6, then n(AUB) is<br>

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Step-by-step explanation:

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Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul

IgorLugansk [536]

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(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: $P(k) = \frac{\lambda^ke^{-k}}{k!}$ where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and $\lambda = 7$ into the formula:

$P(k=7) = \frac{7^7e^{-7}}{7!}$

$P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%$

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

$P(k\leq2) = P(k=0)+P(k=1)+P(k=2)$

$P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}$

$P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}$

$P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%$

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

$P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%$

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2 years ago