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3 years ago

11

A basketball team will play 12 home games during the season. it has played 6 home games so far. of the season's remaining games,

1/3 will be played at home. The equation 1/3g+6=12 can be used to find the the total number of games,g,remaining this season. How many games,g,remain in the season

1 answer:

Arisa [49]3 years ago

3 0

Answer:

18

Step-by-step explanation:

⅓g + 6 = 12

⅓g = 6 Subtracted 6 from each side

g = 18 Multiplied each side by 3

18 games remain in the season.

Check:

⅓ × 18 + 6 = 12

6 + 6 = 12

12 = 12

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8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)

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Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let $D(5, 7) = (x_2, y_2)$

$C(?, ?) = (x_1, y_1)$

$M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})$

Rewrite the equation to find the coordinates of C

$2 = \frac{x_1 + 5}{2}$ and $4 = \frac{y_1 + 7}{2}$

Solve for each:

$2 = \frac{x_1 + 5}{2}$

$2*2 = \frac{x_1 + 5}{2}*2$

$4 = x_1 + 5$

$4 - 5 = x_1 + 5 - 5$

$-1 = x_1$

$x_1 = -1$

$4 = \frac{y_1 + 7}{2}$

$4*2 = \frac{y_1 + 7}{2}*2$

$8 = y_1 + 7$

$8 - 7 = y_1 + 7 - 7$

$1 = y_1$

$y_1 = 1$

Coordinates of endpoint C is (-1, 1)

9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let $A(-2, -9) = (x_2, y_2)$

$B(?, ?) = (x_1, y_1)$

$M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})$

$-4 = \frac{x_1 - 2}{2}$ and $-5 = \frac{y_1 - 9}{2}$

Solve for each:

$-4 = \frac{x_1 - 2}{2}$

$-4*2 = \frac{x_1 - 2}{2}*2$

$-8 = x_1 - 2$

$-8 + 2 = x_1 - 2 + 2$

$-6 = x_1$

$x_1 = -6$

$-5 = \frac{y_1 - 9}{2}$

$-5*2 = \frac{y_1 - 9}{2}*2$

$-10 = y_1 - 9$

$-10 + 9 = y_1 - 9 + 9$

$-1 = y_1$

$y_1 = -1$

Coordinates of endpoint B is (-6, -1)

b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

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tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

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$P(A\ n\ B) = \frac{22}{100}$

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Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

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Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

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P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

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Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

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So, we have:

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$P(B) = \frac{n(B)}{Sample}$

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Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

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And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

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This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

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