geno3141

If Bernard were told the 19th, he would know that the birthday was May 19.

If Bernard were told the 18th, he would know that the birthday was June 18.

The only way that Albert knows that Bernard doesn't know occurs when Albert was told either July or August (thus eliminating May 19 and June 18).

If now Bernard knows, it couldn't be the 14th because this occurs in both July or August.

July has one date left, the 16th, while August has two dates left, the 15th and 17th.

So, if Bernard now knows, it must be because he was told the 16th.

Birthdate -- July 16.

IQR is the inter-quartile range; the distance between the value that is in the 25th percentile and the value that is in the 75th percentile.

Find the value that is 1/4th the way from the bottom to the top;

find the value that is 3/4th the way from the botton to the top (which is 1/4th the way from the top down);

subtract the first value from the second.

This range should hold the middle 50 percent of the values.

Since a_n  =  2n²

--->   a₁  =  2(1)² =  2(1)  =  2

         a₂  =  2(2)² =  2(4)  =  8

         a₃  =  2(3)² =  2(9)  =  18

Can you do a₄ and a₅?

The amount of discount is 20% of $170.00   =>   0.20 x 170.00   =>   $34.00.

The discounted price is $136.00.

Two distinct, non-parallel, lines will intersect in only one point; that one point will be the solution of that system. Lines will be distinct if they either have different y-intercepts, different slopes, or both different y-intercepts and different slopes. This is situation C (different y-intercept) and situation D (different slope and different y-intercept)

To get an infinite number of solutions, you need both equations to reduce to the same equation.

Situtation B:  y - 2  =  x   --->   y  =  x + 2, the same as the original equation; so this situation gives you an infinite number of solutions.

If you have two parallel lines (same slope but different y-intercepts), you will have no solution, because the lines do not intersect. This is situation A -- both have slope = 1, but the y-intercepts are different.

If you factor x² + 10x + 24, you get (x + 6)(x + 4).

So either a = 6 and b = 4 or a = 4 and b = 6.

The possible values of b are 4 and 6.

The smallest 8 digit number is 10,000,000; the largest is 99,000,000.

If you write all of them: 10,000,000 and 10,000,001 and 10,000,002, etc., you will have an arithmetic sequence (you can always get the next one by adding 1 to the previous one).

The formula for the sum of an arithmetic sequence is:  Sum  =  n(f + l)/2

where n = number of terms = 89,000,001,

          f = first term = 10,000,000  

and     l = last term = 99,000,000

So, the sum of all the 8 digit numbers is:  89 000 001(10 000 000 + 99 000 000) / 2

Similarly, for 10 digit numbers.

If the probability that an object is acceptable is 0.70, and it is assumed that the probability of whether or not an object is acceptable or defective is independent from whether or not other objects are wither acceptable or defiectime(that is, whether or not an object is defective does not influence whether or not any other objects are defective), I believe that the best answer is:

0.70 x 0.70 x 0.70  =  0.343  34.3%

If there were only 100 objects to choose from and 70 are know to be accepable, then 70/100 x 69/99 x 68/98 would be the correct way to proceed; however, if this is a production line that may create thousands every hour, then 0.70³ is a better estimate.

The boat that is traveling 34 mi/h will travel 68 miles in 2 hours.

The boat that is travelind 22 mi/h will travel 44 miles in 2 hours.

The angle between is 100°.

The distance between them is the third side of a triangle. Use the law of cosines:

c²  =  a² + b² - 2·a·b·cos(C)

c²  =  68² + 44² - 2·68·44·cos(100)

Can you finish it from here?