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olga2289 [7]
1 year ago
7

need help with this problem drop down1: 6, 10drop down2: 6, 10drop down3: less than (<), greater than (>), less than or eq

ual to, greater than or equal todrop down4: dashed, soliddrop down5: above, below

Mathematics
1 answer:
Kryger [21]1 year ago
3 0

From the information given,

x represents number of small boxes while y represents number of large boxes.

If each small box can hold 6 books and each large box can hold 10 books, it means that the expression for the number of books that x small boxes and y large boxes can hold is

6x + 10y

If James can pack up to 110 books, it means that the number of books that he can pack is less than or equal to 110. The inequality representing this scenario is

6x + 10y ≤ 110

10y ≤ - 6x + 110

Dividing both sides of the equation by 10, we have

y ≤ - 6x/10 + 110/10

y ≤ - 3x/5 + 11

The graph is shown below

The inequality that represents this sitaution is 6x + 10y ≤ 110

The graph of the inequality wil have solid boundary at y = - 3x/5 + 11

and will be shaded below the line.

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A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla
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Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = P(A)= \frac{13}{100}=0.13

The probability that athlete is a basketball player = P(B)= \frac{52}{100}=0.52

The probability that athlete is both basket ball player and  football player = P(A\cap B) = \frac{9}{100}=0.09

We have to find the probability that an athlete chosen is either a football player or a basketball player P(A\cup B).

Now we know that;

P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

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Step-by-step explanation:

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