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andreyandreev [35.5K]
2 years ago
15

In a high school basketball game, a player on the home team makes two free throws. One student asks the student next to her what

he thinks the probability of hitting two free throws in a row is. The student replies, "The probability of him making a free throw is probably about .6, so hitting two free throws is probably about 1.2." Why can this immediately be dismissed as incorrect?
A.
The final number, 1.2, is a fraction, which can never represent a probability.
B.
The probability of making a free throw can never be .6.
C.
The probability of an event happening twice in a row can never be equal.
D.
The final number is greater than 1, which is not a valid probability.
Mathematics
1 answer:
nirvana33 [79]2 years ago
7 0

Answer:

  • D

Step-by-step explanation:

The probability can be expressed as a number between 0 and 1 or a percent value between zero and 100%.

<u>Correct choice is:</u>

  • D. The final number is greater than 1, which is not a valid probability.

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Katyanochek1 [597]

When matrices are equal, corresponding terms are equal.

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If the first and the last term of an arithmetic progression, with common difference
devlian [24]

The number of terms in the given arithmetic sequence is n = 10. Using the given first, last term, and the common difference of the arithmetic sequence, the required value is calculated.

<h3>What is the nth term of an arithmetic sequence?</h3>

The general form of the nth term of an arithmetic sequence is

an = a1 + (n - 1)d

Where,

a1 - first term

n - number of terms in the sequence

d - the common difference

<h3>Calculation:</h3>

The given sequence is an arithmetic sequence.

First term a1 = 1\frac{1}{2} = 3/2

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Common difference d = 1/9

From the general formula,

an = a1 + (n - 1)d

On substituting,

5/2 = 3/2 + (n - 1)1/9

⇒ (n - 1)1/9 = 5/2 - 3/2

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Thus, there are 10 terms in the given arithmetic sequence.

learn more about the arithmetic sequence here:

brainly.com/question/503167

#SPJ1

Disclaimer: The given question in the portal is incorrect. Here is the correct question.

Question: If the first and the last term of an arithmetic progression with a common difference are 1\frac{1}{2}, 2\frac{1}{2} and 1/9 respectively, how many terms has the sequence?

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Please hurry it’s missing
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Answer:

★†\:1 \frac{3}{8}  \div 1 \frac{1}{12} \\  =  \frac{11}{8}  \div  \frac{13}{12}  \\  =  \frac{11}{8}  \times  \frac{12}{13}  \\  =  \frac{11}{2}  \times  \frac{3}{13}  \\  =  \frac{33}{26} =  \boxed{1 \frac{7}{26}}†✓

<h3><u>1 (7/26)</u> is the right answer.</h3>
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