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topjm [15]
3 years ago
9

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays.Find the probability that he did not hit a bounda

ry. ​
Mathematics
1 answer:
Margarita [4]3 years ago
4 0

Answer:

The probability of NOT hitting a boundary is (4/5).

Step-by-step explanation:

Let E: Be the event of hitting a boundary

now, Probability of any event E =  \frac{\textrm{Number of favorable outcomes}}{\textrm{Total number of outcomes}}

Here, number of favorable outcomes = 6

So, P(E)  = \frac{6}{30}  = \frac{1}{5}

⇒Probability of hitting a six is 1/5

Now, P(E)  + P(not E)  = 1

So, P(not hitting a boundary ) = 1  -  P(hitting a boundary)

= 1 - (1/5)  = 4/5

Hence, the probability of NOT hitting a boundary is (4/5).

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Sarah is paid biweekly. Her annual salary is $33,000.
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Answer:

$1269.23

Step-by-step explanation:

Since Sarah is paid biweekly (every 2 weeks), and there are 52 weeks in a year...

52 ÷ 2=26

Sarah is being paid 26 weeks out of the year.

Divide 33,000 by 26 (I only list 4 places after the decimal):

33000 ÷ 26=1269.2307

Round 1269.2307 to the nearest cent (hundredth):

1269.23

3 0
3 years ago
Can someone answer these questions for me?
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6 0
3 years ago
Given f(x)=-1/7√16-x^2 find f^-1(x)
Vera_Pavlovna [14]

Answer:

f^{-1}(x)=\pm \sqrt{49x^2-16}

Not a function.

Step-by-step explanation:

The given function is;

f(x)=-\frac{1}{7}\sqrt{16-x^2}

Let y=-\frac{1}{7}\sqrt{16-x^2}

Interchange x and y;

x=-\frac{1}{7}\sqrt{16-y^2}

Solve for y;

-7x=\sqrt{16-y^2}

Square both sides

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49x^2^2=16-y^2

49x^2^2-16=y^2

y=\pm \sqrt{49x^2-16}

The inverse is

f^{-1}(x)=\pm \sqrt{49x^2-16}

f^{-1}(x)=\pm \sqrt{49x^2-16} is not a function because one x-value maps onto to different y-values.

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3 years ago
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