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fomenos
3 years ago
15

Ratios that are equivalent to 40:28

Mathematics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

10:7

Step-by-step explanation:

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5!(3!)/4!<br><br> all include factorials. what to do?
shtirl [24]
5! Is the same as 5x4x3x2x1 and 4! Is the same as 4x3x2x1 because 4! Is in the denominator it cancels the 4x3x2x1 in the numerator and you are left with 5x(3!) witch is the same a 5x3x2x1. The answer=30
6 0
3 years ago
A line passes through (-2, -3) and (4, -6) what is the slope of the line ?
MrRissso [65]
The slope,  or gradient, of the line is 6/-3 which is -2. The gradient is -2.
8 0
3 years ago
Read 2 more answers
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
Determine the value of base b if (152)b = 0x6A. Please show all steps.
DaniilM [7]

Assuming 0x6A is given in base 16, first convert 152_b and 6A_{16} to a common base, say base 10:

152_b=1\cdot b^2+5\cdot b^1+2\cdot b^0=(b^2+5b+2)_{10}

6A_{16}=6\cdot16^1+10\cdot16^0=106_{10}

Then

b^2+5b+2=106

b^2+5b-104=(b-13)(b+8)=0

\implies b=13

3 0
3 years ago
Read 2 more answers
$149 guitar; 29% discount​
Vikki [24]
Divide number of guitars by 100% and multiply into discount given



149/100 x 29

1.49 x 29

43.21



$43.21 had been removed

Therefore $105.79 is the guitar cost after discount



Hope it was useful ☻


5 0
3 years ago
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