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Igoryamba
3 years ago
5

An exam consists of 50 multiple choice questions. Based on how much you studied, for any given question, you think you have a pr

obability of 0.64 of getting the correct answer. Consider the sampling distribution of the sample proportion of correct questions out of 50. (a) Find the mean and standard error of the sampling distribution of this proportion. Mean = (2 decimal places) Standard error = (3 decimal places) (b) What do you expect for the shape of this sampling distribution? approximately normal because n is large cannot be determined because p is greater than 0.5 cannot be determined because n is large approximately normal because the population is normal (c) If truly p = 0.64, calculate the probability of getting a sample proportion less than 0.60? (correct answers on less than 60% of the questions) Probability = (3 decimal places)
Mathematics
1 answer:
astraxan [27]3 years ago
3 0

Answer: (a) mean = 32.00

(b)Standard error = 0.068

It is expected that the shape will approximately large because n is large.

(c)P(z< 0.60)= 0

Step-by-step explanation:

a. Mean = nP= 0.64× 50 = 32.00(2dp)

P= 0.64

b. Standard error = √ p(1-p)/n

= √0.64(0.36)/50

= 0.068(3dp)

c. Using P(Z<X - U/S)

Where S = standard error

U = mean = 0.64

(c)P(z< 0.60)=

= P( z< 0.60 - 0.64/0.068)

= P( z < -5.88)

= 0.

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Answer:

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Step-by-step explanation:

We can write the following system of equations:

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Adding both equations together, we isolate x and get:

2x=22,\: x=11

Plugging x=11 in any of the equations, we can solve for y:

11+y=6,\: y=-5

Verify that the solution pair (11, -5) works \checkmark

Therefore, the two numbers are \fbox{$-5\:\mathrm{and}\: 11$}.

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The least common factor of 8 and 56, 12 and 30, 16 and 24, and 9 and 15
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The lcf of 8 and 56 is 2, the lcf of 12 and 30 is 2, the lcf of 16 and 24 is 2, and the lcf of 9 and 15 is 3.

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What expressions are equivalent to -56z + 28
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Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b
poizon [28]

The roots of the given polynomials exist  $x=8+\sqrt{10}$, and $x=8-\sqrt{10}$.

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

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Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\

$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\

$&x=8+\sqrt{10}, x=8-\sqrt{10}

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$, and

$x=8-\sqrt{10}$.

To learn more about quadratic equations refer to:

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