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Maurinko [17]
4 years ago
10

A multiple choice test 30 questions, and each question has 5 answer choices (exactly one of which is correct). A student taking

the test guesses randomly on all questions. Using the Normal approximation with continuity correction, determine the approximate probability that the student will get at least as many correct answers as he would expect to get with the random guessing approach.
Mathematics
1 answer:
densk [106]4 years ago
7 0

Answer:

Step-by-step explanation:

Given that a multiple choice test 30 questions, and each question has 5 answer choices (exactly one of which is correct).

When a student taking the test guesses randomly on all questions,p for success in each trial = 1/5 =0.2

As there are two outcomes and each event is independent of the other

X no of correct questions is binomial with n = 30 and p = 0.2

If approximated to normal

mean=np = 15 and Variance = np(1-p) = 4.8

Std dev = 2.191

X is normal (15, 2.191)

His expected value = mean = np

Required prob = P(X>15) = P(X>14.5) using continuity correction

=P(Z>-0.5/2.191) = P(Z>-0.23) =0.5910

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3 years ago
What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*
pickupchik [31]
<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation x^4-41x^2=-400.

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

x^4-41x^2+400=-400+400.

x^4-41x^2+400=0.

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

x^4-25x^2-16x^2+400=0

Making it into two groups, we get

(x^4-25x^2)+(-16x^2+400)=0

Factoring out GCF of each group, we get

x^2(x^2-25)-16(x^2-25)=0

(x^2-25)(x^2-16) =0

Factoring out  (x^2-25) and (x^2-16) separately by difference of the squares identity a^2-b^2=(a-b)(a+b), we get

(x^2-25) = x^2-5^2= (x-5)(x+5) and

x^2-16 = x^2-4^2 =(x-4)(x+4).

Therefore,

(x-5)(x+5)(x-4)(x+4) =0

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>
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3 years ago
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Irina-Kira [14]

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Step-by-step explanation:

6 0
3 years ago
The point A(-8, 6) is translated using T: (x,y) → (x + 5. y - 4). What is the distance from A to A'?
Vsevolod [243]
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Now you must find the distance between these two coordinates. To find the distance you must use the distance formula: √<span>(x2 - x1)^2 + (y2 - y1)^2. Since you now have two points, A and A', plug these into the distance formula.

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3 years ago
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The correct answer is C and D .
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