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n200080 [17]
3 years ago
15

I need help!!!!!!!!!!

Mathematics
1 answer:
Andru [333]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

The points will be on the same line if they have the same slope. Calculate the slope of each pair. If the slopes are the same, then they lie on the same line.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{3-2}{1-4} = \frac{1}{-3} = -\frac{1}{3}

m = \frac{y_2-y_1}{x_2-x_1} = \frac{4-2}{-2-4} = \frac{2}{-6} = -\frac{1}{3}

m = \frac{y_2-y_1}{x_2-x_1} = \frac{4-3}{-2-1} = \frac{1}{-3} = -\frac{1}{3}

The slope is the same.

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Lilly saves 2/9 of her allowance is $18 per week.How much does she save eaxh week?
Veronika [31]

Answer:

$4

Step-by-step explanation:

Here is what you can do.

18 / 9 = 2 times 2 = 4 Why.

Because you have to for each week you need to start with the first week so you can get common denominators 4/18 to 18 cross out the 18’s and you got an answer

3 0
3 years ago
In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u
horsena [70]

Answer:

      \large\boxed{\large\boxed{0.119}}

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

<u>1. Start with the first 11 inspected packages:</u>

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

     C(m,n)=\dfrac{m!}{m!(m-n)!}

      C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

      C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

      C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

         C(7,3)\times C(13,8)

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

        \dfrac{C(7,3)\times C(13,8)}{C(20,11)}

Subsituting:

             \dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}

             =\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885

<u>2. The 12th package</u>

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

<u>3. Finally</u>

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

         0.26818885\times 4/9=0.119

6 0
4 years ago
HELPPP MEEEE hurrryyyy
Zielflug [23.3K]

Answer: 1. B 10(15  12 + 8) = 350

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Step-by-step explanation:

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The correct answer is c
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Step-by-step explanation:

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Now, simplify:

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