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Scilla [17]
3 years ago
15

Julia enjoys shooting paper balls into the wastebasket across her office. She misses the first shot 50% of the time. When she mi

sses on the first shot, she misses the second shot 20% of the time.
Part B: What is the probability of making at least one successful throw? (5 points)
Mathematics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

<u>The probability that Julia makes at least one successful throw is 90%</u>

Step-by-step explanation:

1. Let's review the information provided to us to help Julia to answer the question correctly:

Probability that Julia misses the first shot = 50%

Probability that Julia makes the first shot = 50%

When she misses on the first shot, she misses the second shot 20% of the time. Therefore after missing the first shot, she makes the second shot 80% of the time.

Probability that Julia misses both shots = 50% * 20% = 0.5 * 0.2 = 0.1 = 10%

2. What is the probability of making at least one successful throw?

Probability of making at least one successful throw = 1 - (Probability that Julia misses both shots)

Probability of making at least one successful throw = 100% - 10%

<u>The probability that Julia makes at least one successful throw is 90%</u>

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Given:
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Answer:

10-5\sqrt{2}

Step-by-step explanation:

As per the attached figure, right angled \triangle MDL has an inscribed circle whose center is I.

We have joined the incenter I to the vertices of the \triangle MDL.

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As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10\sqrt{2}

Points P,Q and R are at 90 ^\circ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of  \triangleMIL, \triangleMID and \triangleDIL.

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\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL

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2 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
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Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

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First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

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Putting all the values, we have

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The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

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