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3 years ago

A,B,C, and D are all answer choices

Mathematics

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

Empirical formula

0.34

0.33

A^c = event B or event C

Step-by-step explanation:

A = roommate A wins the game

P(A) = (Rock A and Scissors B) + (Scissors A and paper B) + (paper A and rock B)

P(A) = (0.36*0.53) + (0.32*0.25) + (0.32*0.22) = 0.3412

C = game ends in a tie :

P(C) = (RockA and rockB) + (ScissorsA and ScissorsB) + (ScissorsA and ScissorsB)

P(C) = (0.36*0.22) + (0.32*0.53) + (0.32*0.25) = 0.3288

P(B) = 1 - P(A) - P(C)

P(B) = 1 - 0.3412 - 0.3288

P(B) = 0.33

Complement of event A =event B or event C

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April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers

An experiment consist of rolling 3 fair dice (1 red die, a blue die, and a white die) and recording the number rolled on each di

Setler79 [48]

In 1, there are 6 outcomes for each die, so for three dice, the total combination is 6 x 6 x 6 = 216 outcomes. Hence, the probability of any individual outcome is 1/216 

The outcomes that will add up to 6 are
1+1+4 
1+4+1 
4+1+1 
1+2+3 
1+3+2 
2+1+3 
2+3+1 
3+1+2 
3+2+1 
2+2+2 

Hence the probability is 10/216 

In 3, the minimum sum of the three dice is 3. so we start with this
P(n = 3) 
1+1+1 ; 1/216 

P(n = 4) 
1+1+2 
1+2+1 
2+1+1 ; 3/216 

P(n = 5) 
1+1+3 
1+3+1 
3+1+1 
1+2+2 
2+1+2 
2+2+1; 6/216

The sum in 3 is 10/216 or 5/108

3 0

3 years ago

Complete the proof for the following conjecture.

Ray Of Light [21]

Answer:

Statements Reasons

AC+CD=AD and AB+BD=AD Segment Addition Postulate

AC+CD=AB+BD Transitive/Substitution Property

AC=BD Given

BD+CD=AB+BD Substitution Property

CD=AB Subtraction Property

AB=CD Symmetric Property

Step-by-step explanation:

By segment addition postulate, we can say the following two equations:

AC+CD=AD and AB+BD=AD.

By either substitution/transitive property, you can say AC+CD=AB+BD.

You are given AC=BD, so we use substitution and write AC+CD=AB+AC.

After using subtraction property (subtracting both sides by AC), you obtain CD=AB.

By symmetric property, you may say AB=CD.

So let's write it into the 2 column-proof you have there:

Statements Reasons

AC+CD=AD and AB+BD=AD Segment Addition Postulate

AC+CD=AB+BD Transitive/Substitution Property

AC=BD Given

BD+CD=AB+BD Substitution Property

CD=AB Subtraction Property

AB=CD Symmetric Property

Properties/Postulates used:

Transitive property which says:

If a=b and b=c, then a=c.

Substitution property which says:

If a=b, then b can be substituted(replaced with) for a.

Subtraction property which says:

a=b implies a-c=b-c.

Segment Addition Postulate says:

If you break a segment into two smaller pieces then the measurement of that segment is equal to the sum of the smaller two segments' measurements.

3 0

3 years ago

The height of a cylinder is 8 centimeters. The circumference of the base of the cylinder is 14 pi. What is the measurement of th

timurjin [86]

Answer:

Step-by-step explanation:

3 0

3 years ago

Tim and Tom are trying to earn money to buy a new game system over a 3-month period. Tim saved $45.22 each month. If they need a

stira [4]

D is the answer. 135.66

3 0

3 years ago