Hey please answer asap!

Use the following for the next two questions. A large company is interested in investing in software designed to allow a compute

Dmitriy789 [7]

the answer is 66.43

this is because he is the man and I need more message here so that I can send it the answer is 66.43the answer is 66.43

this is because he is the man and I need more message here so that I can send it the answer is 66.43

8 0

3 years ago

Line L passes through the point (-1, 1) and is PARALLEL to the line shown here. An equation of line L is

Pachacha [2.7K]

The problem cannot be fully solved because you did provide the line on which the line is parallel/ but i can give you the steps on how to solve it. first solve the slope of the parallel line by
m = ( y2 - y1 ) / ( x2 - x1 )
then the slope of that line is equal to slope of line passing pooint ( -1 , 1)
then solve the y intercept of the line using
y = mx + b
where b is the y intercept
then you will have the equation of the line

5 0

3 years ago

Read 2 more answers

Hector took out a small loan of $900 for 15 months. The simple interest rate on the loan was 15%. How much will he pay in intere

Liula [17]

So you have to find 15 percent of 900 which would be 135

3 0

3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

A shipment to a warehouse consists of 2,750 MP3 players. The manager chooses a random sample of 50

Juli2301 [7.4K]

MP3 players in the shipment which are likely to be defective is 330.

Step-by-step explanation:

• Probability is event of an occurrence which is uncertain.

• Probability always lies between 0 to 1.

• The sample of 50 MP3 has 6 defective, to solve it has to be converted.

• Convert percentage in 100 multiply numerator and denominator by 2.

• Result is 6/50 converted to 12/100, bring it to 0.12%.

• 12% is defected in 100% of sample.

• It is found here that 2750 is the population and 12% of 2750.

• 2750 * 0.12 = 330, defective pieces in 2750.

• There are three types of Classical,Empirical or Experimental.

• Classical are ‘n’ number of events find the probability occurrence.

• Empirical or experimental is purely based on events.

5 0

3 years ago