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lorasvet [3.4K]
3 years ago
13

5 - x x + 1 - 2 + 3x x + 1 -

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
8 0

Answer:

Step-by-step explanation: dhfhdfjdfdk

horsena [70]3 years ago
6 0

Answer: 5-x=-5x

x+1=x

2+3x=1/3x

x+1=2

Step-by-step explanation:

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Which expressions are equivalent to (7^-2)*(7^6)
Ierofanga [76]

Answer:

D

Step-by-step explanation:

(7^-2)*(7^6)=7^-2+6

......since the base 7 is the same, when u multiply them, you should add the exponents and keep 7 as it is. That will be 7^4, which in equivalent to ans D(7^2)^2.

8 0
3 years ago
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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.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 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 <em>X</em> = 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 <em>X</em> > 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 <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

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 <em>X</em> ≤ 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
14. Find the value of the function:<br> If f(x) = 4x + 4x - 4, find f(0).<br> f(0) =
n200080 [17]

Answer:

<h2>f(0) = -4</h2>

Step-by-step explanation:

f(x)=4x^2+4x-4\ or\ f^*(x)=4x+4x-4=8x-4\\\\f(0)-\text{put}\ x=0\ \text{to}\ f(x):\\\\f(0)=4(0)^2+4(0)-4=4(0)+0-4=0-4=-4\\\\f^*(0)=8(0)-4=0-4=-4

3 0
3 years ago
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Step-by-step explanation:

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40 common factors multiplication

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3 years ago
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