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Setler79 [48]
3 years ago
14

What is the value of -3 2/3 +1/3

Mathematics
2 answers:
zheka24 [161]3 years ago
5 0

Answer:

-10/3

Step-by-step explanation:

emmainna [20.7K]3 years ago
4 0

Exact Form:

−10/3

Decimal Form:

−3.3...

Mixed Number Form:

−3 1/3

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What is the image of point (4, 3) if the rotation is 90º ? Please and thank you.
Zielflug [23.3K]

Answer:  (-3, 4)


Step-by-step explanation: Given point coordinate (4,3).

We need to rotate the given point about 90º.

<em>In order to find the new coordinates of rotation 90°counterclockwise about the origin, we can apply rule (h, k) ---> (-k,h).</em>

Where (h,k) are the coordinates of original image on axes and (-k,h) are the coordinates of rotated image.


In resulting coordinates of the image first swap the x and y coordinates of the original image and then make the sign opposite of each x-coordinate.

On applying rule (h, k) ---> (-k,h), we get

(4,3)  ---> (-3, 4)

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3 years ago
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The snack table had 48 cookies displayed. After the first round of students came by, the number decreased by 25%. How many cooki
Inessa05 [86]

Answer:

36 cookies left

Step-by-step explanation:

48×.25=12

48-12=36 cookies

6 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

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3 years ago
What algebraic expressions are equivalent to 5x squared - 8
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Answer:

1. 2x^2+3x^2-8 (you can only combine when the variable and exponent are the same).

2. 5x^2-4-4

3. 5x^2(-16)-8

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4 years ago
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