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Tema [17]
3 years ago
15

Which measurement is equivalent to 2.34 m??

Mathematics
1 answer:
rjkz [21]3 years ago
7 0
Metric convertion are neede to know the measurement that equivalent to 2.34 m.
m to cm
2.34 m = 2.34 m ( 100 cm / 1m)= 234 m
m to km
2.34 m = 2.34 m ( 1 km / 1000 km)= 0.00234 km
m to mm
2.34 m = 2.34 m ( 1000 mm / 1 m)= 2340 mm
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I need the answer asap pls! :)
AfilCa [17]

Answer:

18m

Step-by-step explanation:

Since the given traingles are similar

We get

\frac{JK}{MN} =  \frac{KL}{ML}

\frac{JK}{42}  =  \frac{21}{49}

JK =  \frac{21 \times 42}{49}

JK =  \frac{3 \times 7 \times 6 \times 7}{7 \times 7}

JK = 3 \times 6 = 18

JK = 18

Hence the distance across river is 18m

7 0
2 years ago
A golfer keeps track of his score for playing nine holes of golf​ (half a normal golf​ round). His mean score is 8080 with a sta
solniwko [45]

Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and 11\sqrt2 respectively.

Step-by-step explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

     and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y)        [ Variance = (standard deviation)²]

\Rightarrow SD(W)=\sqrt{SD^2(X)+SD^2(Y)}

∴E(W)=E(X)+E(Y) = 80 +80=160

            and

∴SD(W)=\sqrt{SD^2(X)+SD^2(Y)}

              =\sqrt{11^2+11^2}

               =\sqrt{2.11^2}

               =11\sqrt2

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and 11\sqrt2 respectively.

                                               

4 0
3 years ago
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maxonik [38]

Answer:

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c) no, because 62.5% of people who support candidate x are from California, but only 50% of all people in the study are from California.

d) yes, because 62.5% of people who support candidate x are from California, but only 50% of all people in the study are from California.

The answer is D.

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