To add those fractions, you must find a common denominator for all of them. To find this, find a number that is a multiple of all three denominators. Then, multiply the top of the fraction by the same number you multiplied to get the common denominator. Finally you add all of the numerators together and keep the denomiator the same.
Example: 1/2+1/4+1/8
The demominators all are factors of the multiple 8. Therefore, we will use 8. Then, since we multiplied 2 by 4 to get eight for the first fraction, we will multiply the top number by 4 also. The new fraction will look like this: 4/8. After you have done that to all of the fractions, the equation will look like this: 4/8+2/8+1/8. Finally add all of the top numbers together and keep the denominator the same. The answer would be 7/8
To add fractions you want to make sure that the bottom numbers are the same so we have to make them the same, but we have to make sure that the fractions are unchanged so we get the correct answer
so if we find out what number that 9,7, and 5 can all multiply into. we find that 9 times 7 times 5 or 315 is the number that will work
so to get them to all be over 315, we must multiply each fraction by a form of 1/1 or x/x or (number)/(same numbe as on top)
so 7/9 times (35/35)=245/315 1/7 times (45/45)=45/315 3/5 times (56/56)=168/315
we add them togther 245/315+45/315+168/315=(245+45+168)/315=458/315 this simplifies to 1 and 143/315
Figure on the right is simply figure on the left but flipped on its side. we know this because the question tells us the shapes are congruent (which means equal, the same)
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given bywhere Here we're given p=0.10 [ success = defective ] n=3
where
