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kenny6666 [7]
3 years ago
11

Help I’m confused on this

Mathematics
2 answers:
Delvig [45]3 years ago
6 0
Add all the numbers:

36 + 10 + 2 + 20 + 26 + 6 = 100

Count the amount of female:

20 + 26 + 6 = 52

Female out  of total = 52/100

if you have to simplify, its going to be 13/25

Hope this helped☺☺
astra-53 [7]3 years ago
3 0
20+26=46+6=52
I am pretty sure this is correct forgivw me if I’m wrong
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Which function is quadratic?
AveGali [126]

Answer:

g(x) is a quadratic function ⇒ 2

Step-by-step explanation:

  • The quadratic function is the function that has 2 as the greatest power of the variable
  • The form of the quadratic function is f(x) = ax² + bx + c, where a, b, and c are constant

Let us use the information above to solve the question

∵ f(x) = 1.5^{x}

∵ x is the exponent of the base 1.5

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∵ g(x) = 500x² + 345x

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→ That means g(x) is in the form of the quadratic function above

∵ g(x) is in the form of the quadratic function above, where a = 500,

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3 0
3 years ago
Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec
o-na [289]

Answer:

(a) The probability of the event (<em>X</em> > 84) is 0.007.

(b) The probability of the event (<em>X</em> < 64) is 0.483.

Step-by-step explanation:

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...

(a)

Compute the probability of the event (<em>X</em> > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                =1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007

Thus, the probability of the event (<em>X</em> > 84) is 0.007.

(b)

Compute the probability of the event (<em>X</em> < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                =\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483

Thus, the probability of the event (<em>X</em> < 64) is 0.483.

5 0
3 years ago
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