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3 years ago

What was her batting average? help plz

Mathematics

2 answers:

Anarel [89]3 years ago

7 0

Answer:

34%

Step-by-step explanation:

Safarina has 34 out of 99 hits.

Find her battling average.

=> 34 hits

=> 99 hits in all

x = value of percentage

=> 34 = x * 99

=> 34 = 99x

Now, let’s divide both sides with 99

=> 34 / 99 = 99x / 99

=> 0.3434343 = x (0.343434… is a repeating decimals, so need to round off to the nearest hundreds)

=> 0.34 = 34/100

Thus, the percentage of Sarafina’s hit is 34%

Zinaida [17]3 years ago

6 0

Answer: 32.67

I search google

Step-by-step explanation:

I hope this helps, plz mark brainliest

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3 years ago

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Step-by-step explanation:

First of all, see this situation as a cumulative binomial distribution. You have isolated trials with a probability of success. This makes it binomial. The wording of the question "what is the probability of at least half..." makes this cumulative.

There are a few ways to calculate this, and I'm not quite sure which way you're familiar with. I'll show the cumbersome way and use wolfram to make the calculation.

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3 years ago

Suppose that a rectangular picture is surrounded by a frame of uniform width of x

user100 [1]

Answer:

Ok so the area of the frame is 18*12 = 216cm^2

Let’s suppose the width of the frame is x

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i hope i helped!

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3 years ago

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Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

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3 years ago

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valentinak56 [21]

Answer:

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Step-by-step explanation:

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2 years ago

What was her batting average? help plz ​

What was her batting average? help plz