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n200080 [17]
3 years ago
9

Need help with solving/explanation of Geometry question

Mathematics
2 answers:
kenny6666 [7]3 years ago
8 0
Y=90-29
y=61
x=(90+17+61)-180
x=42
since it is a right triangle it should equal 180
they are similar triangles so it should equal the same as ML
KL = 9
Hoochie [10]3 years ago
3 0
Ok. we see that one of the angles is 90 (the square at the bottom left), another one is 17. we see that there is y and there is x. if you look at the picture, y+x= to the third angle. we have to find y first. 
the sum of the angles of any three triangles is 180. lets look at the small  triangle that y is in. the angles are 90, 29, and y. 90+29+y must equal 180. 90+29=119.
119+y=180. minus 119 from both sides to get y=61. now we know that y=62.
now look at the big triangle. we have 90, 17, and y+x. we have 90+17+y+x=180. 90+17=107, so 107+y+x=180. minus 107 from both sides to get y+x=73. we already know that y=62, so 62+x=73. minus 62 on both sides to get x=11.
problem 2. we see that this figure is semetrical. it has a strait line going up from the center. we also see that the base of either half of the triangle is 6, again, proving that this is semetrical. if we fold this figure in half along the line, k will line with m, and the lines LM, and LK will line up. they are the same lenghs, LM=9, so LK=9 if you have any questions regarding this answer, just ask in the comments

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The question is incomplete. Here is the complete question

It was estimated that because of people switching to Metro trains about 33000 of CNG 3300 tons of diesel and 21000 tons of petrol was saved by the end of year 2007 (I) find a fraction of the quantity of diesel saved to the quantity of petrol saved (ii) find the quantity of cng to the quantity of diesel saved

Answer:

(I) 11/70

(II) 1/10

Step-by-step explanation:

Metro trains contain 330,000 tons of CNG

3300 tons of diesel was saved

21000 tons of petrol was saved at the end of 2007

(I) The fraction of the quantity of diesel saved to the quantity of petrol saved can be calculated as follows

= 3300/21000

= 11/70

(II) The fraction of the quantity of CNG saved to the quantity of diesel can be calculated as follows

= 3300/33000

= 1/10

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(Small sample confidence intervals for a population mean) suppose you are taking a sampling of 15 measurements. you find that x=
Luda [366]

Answer:

The 99% confidence interval is  71.67 < \mu < 78.33

Step-by-step explanation:

From the question we are told that

     The sample  size  is  n  =  15

      The  sample  mean is  \= x  =  75

        The  standard deviation is  s =  5

 Given that confidence is  99%  then the level of significance is mathematically represented as

              \alpha  =  100 -  99

             \alpha  =  1\%

             \alpha  =  0.01

Next we obtain the critical values of  \frac{ \alpha }{2} from the normal distribution table

   The  value is

                  Z_{\frac{ \alpha }{2} } = 2.58

Generally the margin for error is mathematically represented as

            E =  Z_{\frac{ \alpha }{2} } *  \frac{ s}{ \sqrt{n} }

=>         E =  2.58  *  \frac{ 5}{ \sqrt{15} }

=>         E =  3.3307

   The  99% confidence interval is mathematically represented as

             \= x  -E  <  \mu  <  \= x  +E

=>          75 -  3.3307 <  \mu

=>          71.67 < \mu < 78.33

3 0
3 years ago
Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proporti
hodyreva [135]

Answer:

The sample size n = 4225

Step-by-step explanation:

We will use maximum error formula = \frac{z_{a} S.D}{\sqrt{n} }

but we will find sample size "n"

\sqrt{n}  = \frac{z_{a} S.D}{maximum error}

Squaring on both sides , we get

n  = (\frac{z_{a} S.D}{maximum error})^2

Given 99% confidence interval (z value) = 2.56

given maximum error = 0.02

n  = (\frac{z_{a} p(1-p)}{maximum error})^2

n≤ (\frac{2.56X\frac{1}{2} }{0.02}) ^{2}    ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 4225

<u>Conclusion</u>:

The sample size of two samples is  n = 4225

<u>verification</u>:-

We will use maximum error formula = \frac{z_{a} S.D}{\sqrt{n} } = \frac{2.56 1/2}{\ \sqrt{4225} }  = 0.0196

substitute all values and simplify we get maximum error is 0.02

4 0
3 years ago
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