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Eva8 [605]
3 years ago
10

Estimate the quotient for 49 / 311

Mathematics
2 answers:
zaharov [31]3 years ago
7 0
If you’re estimating, you’d usually round the numbers and proceed.
49 would round up to 50.
311 would round down to 300.
300 / 50 = 6.
(Unless you were doing 50 / 300, which would be -0.166666...)

Hope that helped! :)
Makovka662 [10]3 years ago
4 0
To get a close estimate, we can round 49 up to 50 and 311 down to 300, obtaining an estimate of 50/300 = 1/6, or 0.1666... as a repeating decimal. That decimal approximation is a little less than one hundredth away from the actual decimal approximation of ≈ 0.1576
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Answer:

The minimum value of C is 20

Step-by-step explanation:

we have

2x+3y\geq 24 ----> constraint A

4x+y\geq 38 ----> constraint B

x\geq 0 ----> constraint C

y\geq 0 ----> constraint D

Using a graphing tool

The solution set of the constraints is the shaded area

see the attached figure

The vertices of the shaded area are (0,38),(9,2) and (12,0)

To determine the minimum value of C substitute the values of x and y of each vertex in the objective function and then compare the results

C=2x+y

For (0,38) ---> C=2(0)+38=38

For (9,2) ---> C=2(9)+2=20

For (12,0) ---> C=2(12)+0=24

therefore

The minimum value of C is 20

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3 years ago
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Step-by-step explanation:

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What is numerical expression?<br> Three less than 11 times the sum of 9-2
sweet-ann [11.9K]

Answer:

11(9 - 2) - 3

Step-by-step explanation:

This is how you write the expression.

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2 years ago
Question 1
fiasKO [112]
The scale factor is 3
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The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability th
neonofarm [45]
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It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function.  Tables and software are available for that purpose.

Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.

The Poisson distribution has a PMF (probability mass function) 
P(k):=\frac{\lambda^ke^{-\lambda}}{k!}
with &lambda; = 1
=>
P(0):=\frac{1^0e^{-1}}{0!}=0.3678794
P(1):=\frac{1^1e^{-1}}{1!}=0.3678794
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=>
P(k
or
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3 0
2 years ago
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