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lilavasa [31]
3 years ago
9

Tiliat is the ratio of rupees 1 and 50 paisa​

Mathematics
1 answer:
dybincka [34]3 years ago
8 0

Answer:

150 paisa means 1 rupees ok 50 paisa

Step-by-step explanation:

left

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if BA is extended all the way through A creating BF and A becomes the midpoint of BF, then what are the coordinates of F
gogolik [260]

Answer:

F(2x_A-x_B,2y_A-y_B)

Step-by-step explanation:

Let points A, B and F have coordinates A(x_A,y_A),  B(x_B,y_B) and F(x_F,y_F).

If BA is extended all the way through A creating BF and A becomes the midpoint of BF, then the midpoint A of the segment BF has coordinates:

\dfrac{x_B+x_F}{2}=x_A\\ \\\dfrac{y_B+y_F}{2}=y_A

Express coordinates of point F:

x_B+x_F=2x_A\Rightarrow x_F=2x_A-x_B\\ \\y_B+y_F=2y_A\Rightarrow y_F=2y_A-y_B

Hence,

F(2x_A-x_B,2y_A-y_B)

3 0
3 years ago
Please help!! I'm confused
olga nikolaevna [1]

nsjsjjsksnndjsndn

Step-by-step explanation:

jsjsjsjsjjsjmsmskzkkennxjzkwmnsks

7 0
3 years ago
Simplify 8(5y-4x)/15 - ( - 3x + 7y/2 +3x - 4y)​
vekshin1

Answer:

that's it I think its clear

7 0
3 years ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
3 years ago
Plz help me with this question
Tems11 [23]
The answer is 1/12, 2/12, 1/4, 1/2
8 0
4 years ago
Read 2 more answers
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