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2 years ago

Use the rule for order of operations to simplify the expression as much as possible 30-2(4*9-26)=

Mathematics

1 answer:

kap26 [50]2 years ago

3 0

20 aahahahahahahHHajajajahaj ajajaja

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What me the answer to s-8=9

Irina-Kira [14]

Answer: -2 i believe because if you take and subtract one from 8 you should get 9

Step-by-step explanation:

5 0

2 years ago

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When randomly choosing two cards from a standard deck of cards without replacement, what is the probability of choosing a face c

Radda [10]

There are 12 face cards and 4 aces to choose from. For the first card, the probability of drawing a face is $\dfrac{\binom{12}1}{\binom{52}1}=\dfrac{12}{52}$ . Then the probability of drawing an ace second is $\dfrac{\binom41}{\binom{51}1}=\dfrac4{51}$ .

So the probability of drawing these two particular cards in the given order is

$\dfrac{12}{52}\cdot\dfrac4{51}=\dfrac4{221}$

4 0

3 years ago

Lucie drank 3/4 of a pint of bottled water. One serving of the water is 7/8 of a pint. How much of a serving did she drink.

attashe74 [19]

7/8
3/4
I'm order for them to have the same denominator you need to multiply so:
¾×2=6/8
So she drank 6/8 pints of water

8 0

2 years ago

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The sum of a number and -3 is equal to -91.Find the number

larisa86 [58]

The sum of -3 and a number equal to -91 is -88

7 0

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:

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

4 0

2 years ago