If 10% of plain M&Ms are blue and you have a sample of 20 M&Ms only two of them will be blue. If you need to have 15 blue M&Ms, then your sample size should be 150 M&Ms because 15 is 10% of 150.
Let M be the mean. P( X > 5.52) =0.1271 Standardize X by subtracting the mean and dividing by the standard deviation. z = (52.2-M) /4 From the normal distribution tables, P( z > 1.14 )=0.1271 Therefore, (52.2-M) / 4 = 1.14 Solve for M, the mean. 52.2-M = 4(1.14) = 4.56 M=52.2-4.56 =47.64
I solve this problem in two steps: Step 1: How much space does one pea take up? Step 2: How much space does 1 trillion peas take up? 1) How much space does one pea take up? I will assume one pea would occupy a square area 5 mm by 5 mm, which equals 25 mm2. 2) A trillion peas, requires me to use scientific notation, 1 trillion = 10^12. So our trillion peas takes up 25 *1012 mm2. Now 1 m = 1000 mm, so 1 m2 = 10^6 mm2, and 1 km = 1000 m, so 1 km2 = 10^6 m2, so 10^12 mm2 = 1 km2. Now, 25 * 1012 mm (1 km2/1012 mm2) = 25 km2 is the area on earth that one trillion peas would cover. Remember: One thousand = 103, One million = 106, one billion = 109, one trillion = 1012 Also, when you find a problem that seems too big to solve, try finding a small problem to solve, which will help you to solve the bigger one.
There are many many ways. One method is called mark-recapture. The simplest method involves taking 2 samples. In the first sample, all the animals that are captured are marked (this can be leg bands, ear tags, toe clipping, or even using photoID in the case of whales/tigers). The size of this sample is called M to denoted that this is now our population of Marked animals. The second sample, which is collected at a later date, will usually (hopefully) contain some of the previously marked animals, and some animals that weren't previously caught. The size of the second sample is denoted n, and the number of marked animals in it is called m. But we want to know N - the total population size. We can assume that the ratio of marked animals in our second sample (m/n) is the same as the ratio of marked animals in the population (M/N). Therefore: M/N=m/n Rearrange it: N=(n*M)/m We know the values for M,n, and m so we can figure out N. This is the simplest case, and is know as the Lincoln-Peterson estimator. There are many extensions to this that allow for more samples etc.
I have some trouble understanding your question. I will answer the following questions: a) What is the area of a square, 500 m on a side? Ans: 500 x 500 = 250,000 m2. b) If I had a square with an area of 500 m2, what is the size of each side? sqrt(500) = 22.3 m
LPP deals with solving problems which are linear . ex: simlpex method, big m method, revised simplex, dual simplex. NLPP deals with non linear equations ex: newton's method, powells method, steepest decent method
Ans. 1. The big M method solves the problem in one pass, whereas the two phase method solve it in two stages. 2. The big M method is computionally inconvenient due to existence of the large number M The two phase method does not involve M during computation 3. The big M method present a difficulty when the problem is solved on digital computer, but there is no such type of problem in two phase method.
The equation given is not enough to solve for k, m, and n as it has 3 unknowns and only 1 equation. You need at least 2 or more equations to solve for the unknowns.
Propeller Immersion : (dA - (xtip X Trim)/Lpp - H) / Dpx100 (in %)Where dA : Draught at A.P (in m)H : Height of lowest tip of propeller bladeDp : Propeller Diameterxtip :x Position of propeller tip from A.P (in m)Lpp : Length Between perpendicular of the ship.================Mosfofa Sowkot ImranNaval ArchitectBangladesh University of Engineering & Technology (BUET)
how big is an m and m
This is a hyperbola. It is best approached using Fermat's factorisation method. Seefermat-s-factorization-methodor google wikepedia. I don't know of any faster approach.
m= -3
m+36
use this method.. d- divide b- bring down m- multiply s- subract
-12=m+4 -16=m
m-12=25. m=37
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