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2 years ago

Joshua owns a fish tank that measures 5.5 m x 2.2m x 0.5m.

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

4 0

6.05 m^3 ; You can check this by simply multiplying all the numbers you provided since volume is l*w*h.

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3 years ago

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Whats the difference between 24 and -14.

liubo4ka [24]

The difference between 24 and -14 is 38

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2 years ago

Express the radical using the imaginary unit, i

Kipish [7]

Answer:

7i.

Step-by-step explanation:

Note : The given expression must be $\sqrt{-49}$ .

We have to write this radical value in the imaginary value.

The given value can be rewritten as

$\sqrt{-49}=\sqrt{49\times (-1)}$

$\sqrt{-49}=\sqrt{49}\times \sqrt{-1}$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$\sqrt{-49}=7\times i$ $[\because \sqrt{-1}=i]$

$\sqrt{-49}=7i$

Therefore, the required expression is 7i.

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3 years ago

On a newly developed IQ test, an individual scores at the 110 level on the first half of the test, and 150 on the second half of

EastWind [94]

Answer:

Reliability

Step-by-step explanation:

The Reliability of a measurement or observation refers to its repeatability. If you measure or observe the same thing twice, how close are the two measurements or observations?

#The test results 110 1nd 150 are between 27% and 36% depending on which direction you take. That variation is wide apart and the test can't be considered reliable.

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3 years ago

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst

grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let X = number of telephone calls.

The average number of calls per minute is, λ = 3.0.

The random variable X follows a Poisson distribution with parameter λ = 3.0.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...$

(a)

Compute the probability of X = 0 as follows:

$P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498$

Thus, the probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of X > 5 as follows:

P (X > 5) = 1 - P (X ≤ 5)

$=1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084$

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of X = 3 as follows:

$P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09$

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of X ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

$=\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578$

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0

3 years ago