Which statement is true about the solution to the two equations? (1 point)

2 answers:

8 0

The first statement is true, because
|3x-1|+7=2 => |3x-1| = -5 (There is no solution. The absolute value is always positive, or zero).
|2x+1| +4=3 => |2x+1|=-1 (There is no solution either, for the same reason).

Lorico [155]3 years ago

6 0

Answer:

the first statement is correct

Step-by-step explanation:

l3x - 1| + 7 = 2 and |2x + 1| + 4 = 3

-Equation 1 and equation 2 have no solutions.

-Equation 1 has no solution, and equation 2 has solutions x = 0, 1.

-The solutions to equation 1 are x = -1.3, 2, and equation 2 has no solution.

-The solutions to equation 1 are x = -1.3, 2, and equation 2 has solutions x = 0, 1.

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2 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:

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