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Vinvika [58]
3 years ago
12

A pet store contains 35 light green parakeets (14 females and 21 males) and 44 sky blue parakeets (28 females and 16 males). You

randomly choose one of the parakeets. What is the probability that it is a female or a sky blue parakeet?
Mathematics
1 answer:
snow_tiger [21]3 years ago
7 0
You can use this formula <span>P(AorB) = P(A) + P(B) - P(AandB) 

Given:
35 LG (14 F & 21 M)
44 SB (28 F & 16 M)

Req:
- the probability that it is a female (F) or a sky blue (SB)

Sol:
</span>P(F or SB) = P(F) + P(SB) - P(F and SB) 
                 = [(14 F + 28 F)/(35 + 44)] + [(44 SB)/(35 + 44)] - [(28 F)/(35 + 44)]
                 = 53.16 + 55.70 - 35.44
                 = 73.42%

You have to deduct 28 female parakeets from 44 sky blue parakeets because the 28 parakeets are already accounted for in the female parakeets. You can also think of how many ways you can choose a female parakeet and a sky blue parakeet. Add all female parakeets (14 + 28) = 42. Sky blue parakeet equaled to 44. Minus the 28 female parakeets included in the sky blue parakeet to avoid double counting. 42 + 44 - 28 = 58 divided by 79 (35 + 44) total parakeets = 73.42%


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The sum of two numbers is 56. One number is 2 more than the other. What is the smaller number?
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Step-by-step explanation:

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Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2
Inessa05 [86]

Answer:

f_{ave} = {\frac{10 }{3\pi }

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3\pi/2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

f_{ave} = \frac{1}{b - a}\int\limits^b_a {f(x)} \, dx

Now,

Here a = 0, b = \frac{3\pi }{2}, f(x) = 5sec^{2}(\frac{x}{6} )

Now,

f_{ave} = \frac{1}{\frac{3\pi }{2}  - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx

      = {\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]

      = {\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]

      = {\frac{10 }{3\pi }}[ {1 - 0 }  ]

      = {\frac{10 }{3\pi }

⇒f_{ave} = {\frac{10 }{3\pi }

3 0
2 years ago
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