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3 years ago

Given the equation 6(8g + 2) + 45 = 153, write a scenario similar to Parts A and B

Mathematics

2 answers:

Zanzabum3 years ago

5 0

Answer:

Step-by-step explanation:

6(8g+2)+45=153

48g+12+45=153

48g+57=153

48g=153-57

48g=96

g=96÷48

g=2

Semenov [28]3 years ago

4 0

Answer:

6(8g + 2) + 45 = 153

48g + 12 + 45 = 153

48g + 57 = 153

- 57 - 57

48g = 96

g = 2

each family bowled 2 games.

Step-by-step explanation:

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Find the slope between the points (−3,−5) and (10,-5)

balandron [24]

Answer:

Step-by-step explanation:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$m = \dfrac{-5 - (-5)}{-3 - 10}$

$m = \dfrac{0}{-13}$

$m = 0$

3 0

2 years ago

Read 2 more answers

Through any 2 points there is exaclty_____line.

mr Goodwill [35]

Answer:

One

Step-by-step explanation:

Through any 2 points, there is exactly one line.

6 0

2 years ago

Read 2 more answers

A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

Eddi Din [679]

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

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

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

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

$D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{1}{455} = 0.0021$

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

$D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{36}{455} = 0.0791$

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

3 0

2 years ago

I conclude from the information given that he is just very lucky. I believe he is lucky, because it is very hard to perform tric

VikaD [51]

Answer:

I believe that the owner of the dice is actually not lucky. There are people who practice throwing dice in a controlled way in order to get a specific side of the dice. If he practices his dice throwing a lot, then he would have a much higher chance of winning at a game of dice.

Step-by-step explanation:

Since you said he was lucky, I decided to make an answer saying that he wasn't lucky so you can choose between both of them. Your answer was good as well.

7 0

2 years ago

If LaTeX: m\angle ABF=8s-6m ∠ A B F = 8 s − 6 and LaTeX: m\angle ABE=2\left(s+11\right)m ∠ A B E = 2 ( s + 11 ), find LaTeX: m\a

kolbaska11 [484]

Answer:

Step-by-step explanation:

Given the angles ∠ABF=8s-6, ∠ABE = 2(s + 11), we are to find the angle ∠EBF. The following expression is true for the three angles;

∠ABF = ∠ABE + ∠EBF

Substituting the given angles into the equation to get the unknown;

8s-6 = 2(s + 11)+ ∠EBF

open the parenthesis

8s-6 = 2s + 22+ ∠EBF

∠EBF = 8s-6-2s-22

collect the like terms

∠EBF = 8s-2s-22-6

∠EBF = 6s-28

factor out the common multiple

∠EBF = 2(3s-14)

Hence the measure of angle ∠EBF is 2(3s-14)

8 0

3 years ago