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murzikaleks [220]
3 years ago
13

5-6. Find the constant of proportionality (unit rate) for each set of values. Then use the constant of proportionality to write

an equation that relates the two values in the table.
5. profit per shirt sold pound Shirts (s) 5 10 15 Profit (p) $7.50 $15.00 $22.50 Apples (a) 4 5 6 Price (p) $7.96 $9.95 $11.94
Hi

6 . Price per

7-8. Determine whether the relationship between the two quantities shown in the table is proportional by graphing on the coordinate plane. Explain your reasoning.
Number of 1 2 3 4 5 Pen sCost $2 54 56 58 $10 7. Cost of Buying Pens Number of 1 2 3 4Minutes Words Typed 50 90 140 180 8. Words Typed.

Solve this fast and I’ll give you 47 points
Mathematics
1 answer:
kakasveta [241]3 years ago
5 0

5. The constant of proportionality is 1.5

The equation is <em>p </em>= 1.5×s

6. The constant of proportionality is 1.99

The equation is p = 1.99 × a

7. The variables Number of Pens and Cost  are not proportional

Please find attached the required graph

8. The variables Number of minutes and Words Typed are not proportional

Please find attached the required graph

The procedure for finding the answers are as follows;

5. The given data are presented as follows;

\begin{array}{ccc}Shirts \ (s)&&Profit \ (p)\\5&&7.50\\10&&15.00\\15&&22.50\end{array}

Where two variables, <em>s</em> and <em>p</em> are proportional, we get;

p ∝ s

Therefore;

p = C × s

C = p/s

Where;

C = The constant of proportionality

Therefore, the constant of proportionality, <em>C</em>, of the given variables, (number of shirts, <em>s</em>, and profit, <em>p</em>, is found as follows;

C = 7.50/5 = 15.00/10 = 22.50/15 = 1.5

The constant of proportionality, C = 1.5

The equation that relates the two values is <em>p </em>= 1.5×s

6.  For the apples to price relationship, we have;

\begin{array}{ccc}Apples \ (a)&&Price\ (p)\\4&&7.96\\5&&9.95\\6&&11.94\end{array}

Therefore;

p ∝ a

p = C × a

C = p/a

Plugging in the values gives;

C = 7.96/4 = 9.95/5 = 11.94/6 = 1.99

The constant of proportionality, C = 1.99

Therefore, the equation relating the two values is p = 1.99 × a

7. The given data is presented in a tabular form as follows;

\begin{array}{ccc}Number \ of \ pens  &&Cost\ \\1&&52\\2&&54\\3&&56\\4&&58\end{array}

A set of data is proportional or has a proportional relationship if their <em>x</em>, and therefore, y-intercept is (0, 0)

From the graph of the data, created with MS Excel, the y-intercept is 50 which is not equal to zero, therefore, the relationship between the data is not a proportional relationship

8. The given data is presented in a tabular form as follows;

\begin{array}{ccc}Number \ of \ Minutes&&Words \ Typed\ \\1&&50\\2&&90\\3&&140\\4&&180\end{array}

From the graph of the data, we have that the y-intercept of the line of best fir is 5, therefore, the relationship is not a proportional relationship

Learn more about proportional relationships here;

brainly.com/question/24289972.

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A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
Hello. I need help on this question. Please answer if you know. Provide the answer with working. Thanks.
11Alexandr11 [23.1K]

Answer:

a) - 11x - 94y / 12    b)  x -4 / 5- 3x   c) 12x - 11 /10y

Step-by-step explanation:

a)  2x - 6y / 4    -  4y - x/ 12    -   4x - 20y /3

6x - 18y / 12 -     4y - x / 12              -  16x - 80y / 12

= 5x - 14y / 12     - 16x - 80y / 12

- 11x - 94y / 12

b) x /5 -    4/ 3x =  x -4 / 5- 3x

c) 2x -3/ 5y  -    5- 2x / 10y + x/y

4x - 6 /10y   -  5-2x/ 10y +  10x / 10y

2x - 11 / 10  + 10x / 10y

= 12x - 11 /10y

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