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prisoha [69]
3 years ago
14

Select the locations on the number line to plot the points 10/2 and −9/2 .

Mathematics
2 answers:
tresset_1 [31]3 years ago
4 0

Answer:

5 and -4.5

Step-by-step explanation:

10/2=5

and -9/2=-4.5

-5  -4.5 -4 -3 -2 -1 0 1  2 3 4 5

Lisa [10]3 years ago
4 0

Answer:

The number line is in the attached picture.

Step-by-step explanation:

We have the points 10/2 and −9/2.

The point 10/2 can be expressed as:

\frac{10}{2} =5

The point -9/2 can be expressed as:

- \frac{9}{2}= - \frac{8+1}{2}= -(\frac{8}{2} + \frac{1}{2})= -(4 + \frac{1}{2} )=-4\frac{1}{2}

So for the first point we have to count 5 places from zero to the right.

For the second point, we have to count 4 places and a half from zero to the left.

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I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.
mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

3 0
3 years ago
In a sociology class there are 1414 sociology majors and 1111 non-sociology majors. 33 students are randomly selected to present
saul85 [17]

Answer: 0.4065

Step-by-step explanation:

Given : In a sociology class there are 14 sociology majors and 11 non-sociology majors.

Total students = 14+11=25

Number of  students are randomly selected = 3

Then, the number of ways to select 3 students from 25 students :-

^{25}C_3=\dfrac{25!}{(25-3)!3!}=\dfrac{25\times24\times23\times22!}{22!3!}\\\\=2300

Number of ways to select  at least 2 of the 3 students re non-sociology majors :-

^{14}C_1\times^{11}C_2+^{14}C_0\times ^{11}C_{3}\\\\=(14)\times\dfrac{11!}{2!(11-2)!}+(1)\times\dfrac{11!}{3!(11-3)!}\\\\=(14)(11\times5)+\dfrac{11\times10\times9}{6}\\\\=770+165=935

The probability that at least 2 of the 3 students selected are non-sociology majors will be :-

\dfrac{935}{2300}=0.40652173913\approx0.4065

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grigory [225]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
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