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SVEN [57.7K]
2 years ago
9

I need help someone please help me please

Mathematics
1 answer:
Butoxors [25]2 years ago
8 0

Answer:

AC = 13.3

Step-by-step explanation:

6+4 = AB

10 / 6 = AC / 8

AC = 13.3

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x = 22/21 , y = 8/3

Step-by-step explanation:

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What is reminder if you devide 786 to 6?​
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There is no reminder as the answer will be an integer 131.

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Find the values of k for which y = kx + 1 is a tangent to the curve y = 2x^2 + x +3
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2 years ago
Reflect the x axis A B C D
Ronch [10]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y ) → (x, - y ) , then

A (- 1, - 17 ) → A' (- 1, 17 )

B (0, - 12 ) → B' (0, 12 )

C (- 5, - 11 ) → C' (- 5, 11 )

D (- 6, - 16 ) → D' (- 6, 16 )

3 0
2 years ago
A box in a supply room contains 18 compact fluorescent lightbulbs, of which 6 are rated 13-watt, 8 are rated 18-watt, and 4 are
V125BC [204]

Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22

b) Probability(each bulb different rating)

= 0.24

Step-by-step explanation:

There are 6 13watts bulbs

8 18watts bulbs

4 23watts bulbs

Total bulbs = 18

a) Probability that all 3 bulbs are 18watts

Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways

Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways

Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12

Number of ways total= 15×12=180waya

Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22

b) Probability P( all 3 bulbs are 1 from each rating)

Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways

Probability = 192/815 =0.24

6 0
2 years ago
Read 2 more answers
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