Find e-0.3 A) -0.74 B) 0.74 C) -1.35

Ramen purchase a box of candy for himself on the way back to the dorm he ate 2/3 of the candy while he was putting away grocerie

Aleksandr [31]

Answer:

12 chocolates

Step-by-step explanation:

Represent the original number of chocolates by c.

Then (1/4)(2/3)c = 2

Solving this for c, we multiply both sides by 4: (2/3)c = 8

and then multiply both sides by (3/2): c = (3/2)(8) = 12

Originally the box contained 12 chocolates.

6 0

3 years ago

PLEASE HELP ME ASAP!! I REALLY NEED HELP :((

solong [7]

Hello!

Question 1
The angles in a triangle add up to 180°

Subtract the known angles from 180 to get x

180 - 105 - 40 = 35

The answer is 35
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Question 2
The angles in a triangle add up to 180°

The box means that angles is 90°

Subtract the know angles from 180 to get x

180 - 65 - 90 = 25

The answer is 25
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Hope this helps!

8 0

3 years ago

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

IgorLugansk [536]

Answer:

(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:

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