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zavuch27 [327]
3 years ago
14

Which statement best describe the sides of a rhombus? *

Mathematics
1 answer:
arsen [322]3 years ago
8 0

Answer:

the option A is a correct ans .

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HELP PLSPLSPLS
ddd [48]

all "atomic" or constituent statements are true

at least one "atomic" or constituent statement is true

3 0
3 years ago
Amber says 2.76/6 gets an answer of 0.19.Is Amber's answer reasonable. (Choose One) A:No,because 0.2x3=0.6 B:No,because 0.2x6=1.
Butoxors [25]

Answer:

No, her answer is unreasonable because 0.2 * 6 = 1.2

Step-by-step explanation:

Here in this question, we are trying to use an approximation to see if what was said to be the result of a calculation is reasonable as an answer.

Looking at her division, 2.76/6 = 0.19

To the nearest tenth , 0.19 is same as 0.2, now multiplying 0.2 by 6, what we have is 1.2 ( of which even 0.19 * 6 is expected to be less than this). we can see that this value is quite far from our expected answer of 2.76 which makes the answer an unreasonable one

Thus, the division 2.76/6 = 0.19 is not a reasonable answer

5 0
3 years ago
Determine the distance between point A(-3,7) and the line y = 1/3x-2.
nika2105 [10]

hmmmm i think there is another part of this

8 0
3 years ago
WHOEVER RESPONDS FIRST WILL GET MARKED BEST. If Kawan paints the visible outside faces of her shed what is the total surface are
DIA [1.3K]

the total area will be the sum of the area of the four sides of the shed. In other words it is the curved surface area

3 0
2 years ago
Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four
MatroZZZ [7]

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = 6+4=10

Total number of ways to make a committee of five people from 10 persons :-

^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210

a) Number of ways to make committee that has exactly four women :

^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60

The  probability that committee has exactly four women :

\dfrac{60}{210}=\dfrac{2}{7}

b) Number of ways to make committee that has at-least four women :

^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66

The probability that committee at-least four women :

\dfrac{66}{210}=\dfrac{22}{70}

c) Number of ways that committee has more than 4 women :-

^6C_5\times^4C_0=6

The probability that committee has more than 4 women :-

\dfrac{6}{210}

Now, the  probability that committee has at most four women :-

1-\dfrac{6}{210}=\dfrac{204}{210}=\dfrac{102}{105}

5 0
3 years ago
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