HELP PLEASE Which graph represents the system of inequalities?

Mathematics

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

The answer to your question is in the picture below

Step-by-step explanation:

y + 3/4x < 2

y+ 3/4x = 2

y = -3/4x +2

Interceptions (0,2) and (8/3, 0) It's a dotted line and we are interested in the area below the line

y - 1 ≥ 2x

y - 1 = 2x

y = 2x + 1 Interception, it's not a dotted line with interceptions (0,1) and (-1/2, 0), we are interested in the area up the line.

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is the statement below always, sometimes or never true. Gi at least 2 examples to support your reasoning. The lcm of two numbers

ASHA 777 [7]

The statement is not always true.
For Example...
The LCM of 6 and 8 is not the product of the two numbers
6*8= 48
48 is not the LEAST common multiple
Multiples of 6 are...
6,12,18,24,30,36,42,48...
Multiples of 8 are....
8,16,24,32,40,48...
The least common multiple is 24 not 48 even though they are both common multiples.

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3 years ago

Find the length of AC if AB = 16 and BC = 12.

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AC = 28

Step-by-step explanation:

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3 years ago

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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]

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:

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