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Nadya [2.5K]
3 years ago
5

Find a,b,c and d?..........

Mathematics
1 answer:
cupoosta [38]3 years ago
4 0

9514 1404 393

Answer:

  a = 6, b = 12, c = 6, d = 6√3

Step-by-step explanation:

The two triangles are the "special" triangles.

The 45-45-90 triangle has side lengths in the ratios 1 : 1 : √2.

The 30-60-90 triangle has side lengths in the ratios 1 : √3 : 2.

Then ...

  c : a : 6√2 = 1 : 1 : √2   ⇒   c = a = 6

and ...

  a : d : b = 1 : √3 : 2   ⇒   d = 6√3 and b = 12 . . . (since a=6)

The lengths are ...

  a = 6, b = 12, c = 6, d = 6√3

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Suppose that F(x) = x^3 and G(x) = (4x)^3 − 5. Which statement best compares the graphs of F(x) and G(x)?
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3 years ago
What is z? <br> will give brainliest:)
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What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United Stat
neonofarm [45]

Answer:

\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8

s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3

Sd(X) = \sqrt{118.273}=10.9

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

For this case we can calculate the properties required with the following table:

Interval     Mid point (x)     f           x*f         x^2 *f

_________________________________________

1-10               5.5               40        220         1210

11-20             15.5              15        232.5       3603.75

21-30            25.5             23       586.5       14955.75

>31                35.5             10        355          12602.5

________________________________________

Total                                 88        1394         32372

We assume that the mid point for the class >31 is 35.5 using the problem information.

For this case the expected value would be given by:

\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8

The variance owuld be given by this formula"

s^2 = \frac{n(\sum x^2 f) -(\sum xf)^2}{n(n-1}

And if we replace we got:

s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3

The standard deviation would be just the square root of the variance:

Sd(X) = \sqrt{118.273}=10.9

4 0
3 years ago
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