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DaniilM [7]
3 years ago
5

Is sqrt(25a^2/4) the same as sqrt(25a^2)/sqrt(4)?

Mathematics
2 answers:
Evgesh-ka [11]3 years ago
6 0
LHS\\ \\ =\sqrt { \frac { 25{ a }^{ 2 } }{ 4 }  } \\ \\ ={ \left( \frac { 25{ a }^{ 2 } }{ 4 }  \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ =\frac { { \left( 25{ a }^{ 2 } \right)  }^{ \frac { 1 }{ 2 }  } }{ { 4 }^{ \frac { 1 }{ 2 }  } } \\ \\ =\frac { \sqrt { 25{ a }^{ 2 } }  }{ \sqrt { 4 }  } \\ \\ =RHS
xenn [34]3 years ago
5 0
Yes, because \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt a}{\sqrt b}
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3 years ago
A baseball player has a batting average of 0.25. What is the probability that he has exactly 2 hits in his next 7 at bats
marissa [1.9K]

Answer:

The probability that he has exactly 2 hits in his next 7 at-bats is 0.3115.

Step-by-step explanation:

We are given that a baseball player has a batting average of 0.25 and we have to find the probability that he has exactly 2 hits in his next 7 at-bats.

Let X = <u><em>Number of hits made by a baseball player</em></u>

The above situation can be represented through binomial distribution;

P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r}; x = 0,1,2,......

where, n = number of trials (samples) taken = 7 at-bats

            r = number of success = exactly 2 hits

            p = probability of success which in our question is batting average

                   of a baseball player, i.e; p = 0.25

SO, X ~ Binom(n = 7, p = 0.25)

Now, the probability that he has exactly 2 hits in his next 7 at-bats is given by = P(X = 2)

          P(X = 2) =  \binom{7}{2}\times 0.25^{2} \times (1-0.25)^{7-2}

                        =  21 \times 0.25^{2} \times 0.75^{5}

                        =  <u>0.3115</u>

5 0
3 years ago
Plz answer!!!!!!!!!!!!!!​
makkiz [27]

Answer:

7

Step-by-step explanation:

i know

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Is john 16?
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