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Kobotan [32]
3 years ago
9

Please help me with this question, image attached.

Mathematics
1 answer:
disa [49]3 years ago
8 0
I’d have to say 8 cm
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Kelly took a survey of two different sets of
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Step-by-step explanation:

legit hindi ko alam sagot

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3 years ago
Find the value of 294 cm^2 in m^2
swat32
A meter m is 100 cm.
So m^2 = 100^2 cm^2
Then cm^2 = (m^2)/100^2 =m^2/10000
294cm^2 = 294/10000 m^2 or 0.00294 m^2
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I am confused about the order to go about doing this problem because I tried multiplying first and then dividing but my answer w
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50÷(-25)x(4)
50/-25= -2
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VERY VERY VERY VERY VERY URGENT!!?!!???<br><br><br> These triangles are similar.<br> True<br> False
Deffense [45]

Answer:

True

Step-by-step explanation:

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3 years ago
Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his sc
aivan3 [116]

1) 4.55

2) Short hit

Step-by-step explanation:

1)

The table containing the score and the relative probability of each score is:

Score 3 4 5 6 7

Probability 0.15 0.40 0.25 0.15 0.05

Here we call

X = Miguel's score on the Water Hole

The expected value of a certain variable X is given by:

E(X)=\sum x_i p_i

where

x_i are all the possible values that the variable X can take

p_i is the probability that X=x_i

Therefore in this problem, the expected value of MIguel's score is given by:

E(X)=3\cdot 0.15 + 4\cdot 0.40 + 5\cdot 0.25 + 6\cdot 0.15 + 7\cdot 0.05=4.55

2)

In this problem, we call:

X = Miguel's score on the Water Hole

Here we have that:

- If the long hit is successfull, the expected value of X is

E(X)=4.2

- Instead, if the long hit fails, the expected value of X is

E(X)=5.4

Here we also know that the probability of a successfull long hit is

p(L)=0.4

Which means that the probabilty of an unsuccessfull long hit is

p(L^c)=1-p(L)=1-0.4=0.6

Therefore, the expected value of X if Miguel chooses the long hit approach is:

E(X)=p(L)\cdot 4.2 + p(L^C)\cdot 5.4 = 0.4\cdot 4.2 + 0.6\cdot 5.4 =4.92

In part 1) of the problem, we saw that the expected value for the short hit was instead

E(X)=4.55

Since the expected value for X is lower (=better) for the short hit approach, we can say that the short hit approach is better.

8 0
3 years ago
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