Archive for March, 2015

A manufacturer has a steady annual demand for 22,638

A manufacturer has a steady annual demand for 22,638 cases of sugar. It costs $7 to store 1 case for 1 year, $33 in set up cost to produce each batch, and $20 to produce each case. Find the number of cases per batch that should be produced to minimize cost.

 

Solution:-

The manufacturer should produce 512 cases per batch.

Problem 651

The time required to finish a test is normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the probability that a student will finish the test in less than 70 minutes?

 

Solution:-

84%

Problem 650

The time required to finish a test is normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the z-Score for a student who finishes the test in 45 minutes?

 

Solution:-

 

z = \frac{(45 - 60)}{10}

z =\frac{ -15}{10}

z = -1.5

 

Problem 649

Find the sum of the following infinite geometric series, if it exists.

 \frac{2}{5}+\frac{12}{25}+\frac{72}{125}+...

 

Solution:-

 

S =\frac{ a}{1-r} , where r = common ratio (r ≤ 1) and a = first term.

a = \frac{2}{5}, r = \frac{6}{125}

S = \frac{\frac{2}{5}}{\left ( 1-\frac{6}{125} \right )}

S =\frac{ 50}{119}

So the sum of the infinite geometric series 50/119

Problem 648

Find the sum of the following infinite geometric series, if it exists.

 \frac{1}{2}+\left ( \frac{-1}{4} \right )+\frac{1}{8}+\left ( \frac{-1}{16}  \right )+...

 

Solution:-

 

S = \frac{a}{1-r} , where r = common ratio (r ≤ 1) and a = first term.

a = \frac{1}{2} , r = -1/8

S =  \frac{\frac{1}{2}}{\left ( 1-\left ( \frac{-1}{8} \right ) \right )}

S = \frac{4}{9}

So the sum of the infinite geometric series 4/9
 

Problem 647

Find S15 for 1 + 1.5 + 2.25 + 3.375 + …

 

Solution:-

 

Sn =\frac{ a1(1 - r^{n})}{(1 - r) }, Where a1 = first term , r = common ratio and n = finding term.

a1 = 1 , r = 1.5 , n = 15

S15 =\frac{ 1(1 - 1.5^{15})}{(1-1.5)}

S15 = 873.79

So the value of S15 873.79

 

Problem 646

Find S11 for 1 + 2 + 4 + 8 + …

 

Solution:-

Sn =

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r^{n})}{(1 - r)} , Where a1 = first term , r = common ratio and n = finding term.

a1 = 1 , r = 2 , n = 11

S11 = \frac{1(1 - 2^{11})}{(1-2)}

S11 = 2047

So the value of S11 2047

Problem 645

Find the three geometric means between\frac{1}{2} and 8.

 

Solution:-

 

 

t1=\frac{1}{2} , t5 = 8
 
t5 = r^{(n-1)}
 
8= \frac{1}{2}* r^{5-1}
 
8*2 = r^{4}
 
16 = r^{4}
 
r = 2
 
In geometric sequence we have to find the next number by multiply last number to common ratio.
\frac{1}{2} *2 = 1
 
1*2 = 2
 
2 * 2 = 4
 
So the three geometric means 1, 2, 4

Problem 644

Find the three geometric means between 2 and 162.

 

Solution:-

 

t1=2
t5=162
162 =2* r^{5-1}
81 = r^{4}
r = 3
In geometric sequence we have to find the next number by multiply last number to common ratio.
2*3 = 6
6*3 =18
18*3 = 54

So the three geometric means 6, 18, 54

Problem 643

Find the sixth term of a geometric sequence with t5 = 24 and t8 = 3.

 

Solution:-

 

t5 = 24 =  a*r^{4}

t8 = 3 =  a*r^{7}

Dividing one by the other

r^{3} =\frac{ 1}{8}

r = \frac{1}{2}

t6 = a*r^{5} = t5*r = 24 *\frac{ 1}{2} = 12.

So the value of sixth term 12