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3 years ago

The number of goldfish that can live in a small tank is at most 6

Mathematics

1 answer:

Gelneren [198K]3 years ago

5 0

Answer:

n ≤ 6

Step-by-step explanation:

Representing the number of goldfish that can live in a small tank as n, we can express this constraint as

n ≤ 6

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How many students in all played a game or read

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3 years ago

If cos 0 = 1/2, find the degree measure of 0.

Leokris [45]

Answer:

60°

Step-by-step explanation:

if you look on a unit circle the coordinates correspond to (cos, sin) so at 60° they are (1/2, √3/2)

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3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

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4 years ago

WILL GIVE BRAINLIEST ASAP In the figure below, m Z2 =55° and m 2 KJL = 154º. Find m 21.

Elan Coil [88]

Answer:

99°

Explanation:

given:

m∠2 = 55°

solve:

m∠1 + m∠2 = ∠KJL

m∠1 = ∠KJL - m∠2

m∠1 = 154° - 55°

m∠1 = 99°

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2 years ago

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Evaluate the expression for a = 3, b = 12 , and c = 4 . 8ab/2bc Enter your answer in the box.

antoniya [11.8K]

To get the answer the first thing you wanna do is plug in the numbers so that would be (8)(3)(12)/(2)(12)(4)= than you would multiply the numbers and you should get 288 divided by 96 and the answer is 3

7 0

3 years ago

Read 2 more answers