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slamgirl [31]
3 years ago
9

A test has multiple choice questions with 5 choices for each answer, only one answer is correct for each question. Suppose a stu

dent guesses the answer to each question. Assuming the guesses are independent, find the probability that the student will not guess correctly on any one question.
Mathematics
1 answer:
Brrunno [24]3 years ago
6 0

Answer:

The probability that the student will not guess correctly on any one question is 4/5

Step-by-step explanation:

The question posed is an application of the binomial probability distribution.

We have been informed that the test has multiple choice questions with 5 choices for each answer that only one answer is correct for each question.

If the student guesses the answer to each question, then the probability that the student will guess correctly on any one question is 1/5.

Therefore, the probability that the student will not guess correctly on any one question is;

1 - 1/5 = 4/5

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Is 16to 14 and 64to60 equivalent
stiks02 [169]
NO. They are not.

We can prove this by turning those numbers into fractions:

16 / 14 = 1.1428  ;  64 / 60 = 1.0667

Or simply:

64 / 16 = 4
60 / 14 = 4.29

To get the equivalent of 16 to 14; we must multiply both numbers by 4.

16 * 4 = 64
14 * 4 = 56

The equivalent of 16 to 14 is 64 to 56.

3 0
3 years ago
What's the circumference of a circle with a diameter of 3 inches? Use 3.14 for pi
DedPeter [7]

Answer:

C≈18.85

Step-by-step explanation:

3 0
2 years ago
When x=-2 and y=-3. - x^2? - 3xy + 2y^3 is
Angelina_Jolie [31]

Answer:

-66

Step-by-step explanation:

-(-2)^2=-4

-3(-2)(-3)+2(-3)^3= -62

-4-62= -66

3 0
3 years ago
Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0
3 years ago
Eric earned 60 points on his last test if there were 75 possible points what was his percent grade
Andru [333]
So 75 points is your 100%. You can do 60 over 75 is equal to X over 100. You divide your numerator(60) by your denominator(75) to get 85.7142857%, or to round to your nearest percent, your answer would be roughly 86%.
6 0
3 years ago
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