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3 years ago

15

Sue has 18 CDs and 30 DVDs. He wants to place them onto the greatest number of shelves so that each shelf has the same number of

CDs and the same number of DVDs. How many shelves will he use?

1 answer:

frez [133]3 years ago

8 0

Cds 3 dvds 5

Sue will use 6 shelves with 3 cds on each and 5 dvds on each.

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can someone please help I don’t get it and I just want answers I have been trying to solve this for 1 hour now

dusya [7]

Answer:

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3. 7 + 4p-5+p+2q = 2 + 5p + 2q

Step-by-step explanation:

basically you can add terms that have the same variable

integers can be added together, Xs can be added, Zs, Ys, As, Bs, Cs, you get the point

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3. 7 + 4p-5+p+2q = 2 + 5p + 2q (do not add different variables p and q ) together

try 4-6 on your own to get this skill down, if you need help with those just let me know

7 0

3 years ago

Denise has $78.22. she wants to buy a computer that cost $29.99. about how much money will denise has left

blsea [12.9K]

Denise has $50 left.

8 0

3 years ago

In order to raise money for holiday gifts for your family, you are planning on starting a dog walking and car wash business. You

kramer

Answer:

Part 1) see the explanation

Part 2) see the explanation

Part 3) see the explanation

Part 4) The graph in the attached figure

Part 5) see the explanation

Step-by-step explanation:

Part 1) Assign a variable to represent the number of hours that you will spend dog walking in November. Write an expression to represent the amount of money you need to earn while dog walking

Let

x ----> the number of hours spent dog walking in November

so

$12x=600$

solve for x

$x=50\ hours$

Part 2) Assign a variable to represent the number of hours that you will spend washing cars in November. Write an expression to represent the amount of money you need to earn while washing cars.

Let

y ----> the number of hours spent washing cars in November

so

$18y=600$

solve for y

$y=33.33\ hours$

Part 3) Write an algebraic model using inequalities that represents the total amount of money earned by dog walking and washing cars during the month of November.

Let

x ----> the number of hours spent dog walking in November

y ----> the number of hours spent washing cars in November

we know that

You need to earn at least $600 during the month of November

The word "at least" means " greater than or equal to"

so

$12x+18y\geq 600$

Part 4) Graph the algebraic model in the first quadrant only.

we have

$12x+18y\geq 600$

using a graphing tool

The solution is the shaded area

see the attached figure N 1

Part 5) Use the graph and algebraic model to answer the following:

a) Why does the graph exist only in the first quadrant?

Because the number of hours cannot be a negative number

b) Are you able to earn exactly $600? Use the solutions of the system to find possible combinations of outcomes that equal exactly $600. Where do all of the combinations occur in the graph?

Yes its possible to earn exactly $600. All the combinations of outcomes that equal exactly $600 occur in the solid line 12x+18y=600

c) Is it possible to earn more than $600? Use the solutions of the system to find possible combinations of outcomes that are greater than $600. Where do all of the combinations occur in the graph?

Yes its possible to earn more than $600. All the combinations of outcomes that are greater than $600 occur above the solid line 12x+18y=600

d) If you work for 10 hours walking dogs and 10 hours washing cars, will you have earned enough money for the holiday gifts?

we have the ordered pair (10,10)

substitute in the inequality

$12(10)+18(10)\geq 600$

$300\geq 600$ ----> is not true

therefore

You must not have earned enough money for holiday gifts.

e) Where does 10 hours walking dogs and 10 hours washing cars fall on the graph? Is this location representative of the solution to the algebraic model?

The ordered pair (10,10) fall below the solid line 12x+18y=600

This location is not representative of the solution to the algebraic model, because the solution of the model is above the solid line

f) How would the algebraic model be different if you needed to earn more than $600? Adjust your algebraic model to show that you must earn more than $600. Would the graph of the model be different from the original? Would you include the line in the solution?

we know that

$12x+18y=600$ ----> represent a linear equation that show the possible combinations of outcomes that equal exactly $600.

The solution is the set of ordered pairs belong to the solid line

$12x+18y\geq 600$ ----> represent a inequality that show the possible combinations of outcomes that are greater than or equal to $600. The solid line is included in this solution

The solution is the shaded area above the solid line (The solid line is included in the shaded area)

g) In complete sentences, explain the difference between a solid line and a dashed line when graphing an inequality. When graphing the two algebraic models, how did you determine which type of line to use?

we know that

When the inequality is of the form ≥ or ≤ we have a solid line

When the inequality is of the form > or < we have a dashed line

The difference is that in the solid line the line is included in the solution and in the dashed line the line is not include in the solution

h) How did you determine which part of the graph of the inequality to shade? What does the shaded area tell you? What does the area that is not shaded tell you?

we know that

when you have an inequality of the form

y> ax+b or y≥ax+b

The shaded area is above the line

when you have an inequality of the form

y< ax+b or y≤ax+b

The shaded area is below the line

The shaded area is the solution set of the inequality (ordered pairs that satisfy the inequality)

The area that is not shaded are the ordered pairs that are not solutions of the inequality (ordered pairs that satisfy the inequality)

5 0

3 years ago

Leon wants to estimate the proportion of the seniors at his high school who like to watch football. He interviews a simple rando

cupoosta [38]

Answer:

Yes, the random conditions are met

Step-by-step explanation:

From the question, np^ = 32 and n(1 − p^) = 18.

Thus, we can say that:Yes, the random condition for finding confidence intervals is met because the values of np^ and n(1 − p^) are greater than 10.

Also, Yes, the random condition for finding confidence intervals is met because the sample size is greater than 30.

Confidence interval approach is valid if;

1) sample is a simple random sample

2) sample size is sufficiently large, which means that it includes at least 10 successes and 10 failures. In general a sample size of 30 is considered sufficient.

These two conditions are met by the sample described in the question.

So, Yes, the random conditions are met.

5 0

3 years ago

A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable

Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 14, p = 0.3$ .

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066$

$P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014$

$P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002$

$P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024$

$P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002$

$P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0$

$P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082$

0.0082 = 0.82% probability that he will pass

6 0

3 years ago