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Lorico [155]
3 years ago
7

Find the difference between the result of one sixth of product of 10 and 3 multiplied by 4 and 30

Mathematics
1 answer:
kherson [118]3 years ago
6 0

Answer:

10 or -10,

Read the explanation

Step-by-step explanation:

Rewrite this problem as a numerical expression. As per the wording of this problem, there can be two expressions derived.

1.     ((\frac{1}{6}(10*3)*4)-30

2.     30-((\frac{1}{6}(10*3)*4)

Simplify, remember the order of operations. The order of operations is the sequence by which one is supposed to perform operations in a numerical expression. This order is the following:

1. Parenthesis

2. Exponents

3. Multiplication or division

4. Addition or Subtraction

Use this sequence when simplifying and solving the expression:

Expression 1

((\frac{1}{6}(10*3)*4)-30\\\\=((\frac{1}{6}(30)*4)-30\\\\=(5*4)-30\\\\=20 - 30\\\\= -10

Expression 2

30-((\frac{1}{6}(10*3)*4)\\\\=30-((\frac{1}{6}(30)*4)\\\\=30-(5*4)\\\\= 30-20\\\\= 10

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At a football game, 65% of people attending were supporting the home team, while 35% were supporting the visiting team. If 1300
Misha Larkins [42]

Answer: 2000

Step-by-step explanation:

At a football game, 65% of people that will attend are supporting the home team, while 35% are supporting the visiting team. 1300 people that attended supported the home team.

To get the total number of people that attended the game, we have to find the number of away supporters as well.

Let the number of people that attended the game be y.

Therefore 65% of y = 1300

65/100 × y = 1300

0.65 × y = 1300

0.65y = 1300

Divide both side by 0.65

0.65y/0.65 = 1300/0.65

y = 2000

The total number of people attending the game is 2000.

8 0
3 years ago
What is the probability that a fair die never comes up an even number when it is rolled six times? (note: enter the value of pro
seropon [69]
Let us compute first the probability of ending up an odd number when rolling a dice. A dice has faces with numbers 1 up to 6. The odd numbers within that is 3 (1, 3 and 5). Therefore, each dice has a probability of 3/6 or 1/2. Then, you use the repeated trials formula:

Probability = n!/r!(n-r)! * p^r * q^(n-r), where n is the number of tries (n=6), r is the number tries where you get an even number (r=0), p is the probability of having an even face and q is the probability of having an odd face.

Probability = 6!/0!(6!) * (1/2)^0 * (1/2)^6
Probability = 1/64

Therefore, the probability is 1/64 or 1.56%.
6 0
3 years ago
20 POINTS
noname [10]

Answer:

EXAMPLE

Pete bought a shirt on sale for $

18

18, which is one-half the original price. What was the original price of the shirt?

Solution:

Step 1. Read the problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don’t understand, look them up in a dictionary or on the Internet.

In this problem, do you understand what is being discussed? Do you understand every word?

Step 2. Identify what you are looking for. It’s hard to find something if you are not sure what it is! Read the problem again and look for words that tell you what you are looking for!

In this problem, the words “what was the original price of the shirt” tell you what you are looking for: the original price of the shirt.

Step 3. Name what you are looking for. Choose a variable to represent that quantity. You can use any letter for the variable, but it may help to choose one that helps you remember what it represents.

Let

p

=

p= the original price of the shirt

Step 4. Translate into an equation. It may help to first restate the problem in one sentence, with all the important information. Then translate the sentence into an equation.

The top line reads:

Step 5. Solve the equation using good algebra techniques. Even if you know the answer right away, using algebra will better prepare you to solve problems that do not have obvious answers.

Write the equation.

18

=

1

2

p

18=12p

Multiply both sides by 2.

2

⋅

18

=

2

⋅

1

2

p

2⋅18=2⋅1

Step 3. Name what you are looking for. Choose a variable to represent that quantity. You can use any letter for the variable, but it may help to choose one that helps you remember what it represents.

Let

p

=

p= the original price of the shirt

Step 4. Translate into an equation. It may help to first restate the problem in one sentence, with all the important information. Then translate the sentence into an equation.

The top line reads:

Step 5. Solve the equation using good algebra techniques. Even if you know the answer right away, using algebra will better prepare you to solve problems that do not have obvious answers.

Write the equation.

18

=

1

2

p

18=12p

Multiply both sides by 2.

2

⋅

18

=

2

⋅

1

2

p

2⋅18=2⋅12p

Simplify.

36

=

p

36=pStep 6. Check the answer in the problem and make sure it makes sense.

We found that

p

=

36

p=36, which means the original price was

$36

$36. Does

$36

$36 make sense in the problem? Yes, because

18

18 is one-half of

36

36, and the shirt was on sale at half the original price.Step 7. Answer the question with a complete sentence.

The problem asked “What was the original price of the shirt?” The answer to the question is: “The original price of the shirt was

$36

$36.”

If this were a homework exercise, our work might look like this:

The top reads,

TRY IT

Step-by-step explanation:

<em>#CarryOnLearning</em>

4 0
2 years ago
Which logarithmic equation is equivalent to 3^2 = 9?
noname [10]

Answer:  log_39=2 is the logarithmic equation which is equivalent to3^2=9.


Step-by-step explanation:

According to the laws of logarithms

log_b\ x=n ,where b=base ,n,x all are positive

It can be written in exponential form such as

x=b^n

Now in the given question x=9 ,b=3 and n=2 such that it will equals to

log_39=2

Hence the logarithmic equation for 3^2=9 islog_39=2.

8 0
3 years ago
Read 2 more answers
Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
3 years ago
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