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1 year ago

Solve for x in the equation ]x^{2} - 12x + 59 = 0

Mathematics

2 answers:

Vedmedyk [2.9K]1 year ago

8 0

Step-by-step explanation:

$x^{2} - 12x + 59 = 0$

Given equaiton is in the form of ax^2 +bx+c=0

we apply quadratic formula to solve for x

$x= \frac{-b+-\sqrt{b^2-4ac} }{2a}$

a= 1 b = -12 and c= 59

$x= \frac{12+-\sqrt{(-12)^2-4(1)(59)}}{2(1)}$

$x= \frac{12+-\sqrt{92}}{2}$

$x= \frac{12+-2\sqrt{23}}{2}$

Divide the 12 and square root terms by 2

$x=6+-\sqrt{23}$

so $x=6+\sqrt{23}$ and $x=6-\sqrt{23}$

Triss [41]1 year ago

3 0

Answer:

Solution for x:

$6+\sqrt{23}i$ and $6-\sqrt{23}i$

Step-by-step explanation:

Given: $x^2-12x+59=0$

It is quadratic equation and to solve for x.

using quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

where, a=1, b=-12 and c=59

Put the values into the formula

$x=\dfrac{12\pm\sqrt{12^2-4\cdot 1\cdot -59}}{2(1)}$

$x=\dfrac{12\pm \sqrt{-92}}{2}$

As we know, $\sqrt{-1}=i$

$x=6\pm\sqrt{23}i$

Exact value of x : $6+\sqrt{23}i$ and $6-\sqrt{23}i$

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the distribution of scores on a recent test closely followed a normal distribution wotb a mean of 22 and a standard deviation of

soldi70 [24.7K]

Answer:

1) 22.66%

2) 20

Step-by-step explanation:

The scores of a test are normally distributed.

Mean of the test scores = u = 22

Standard Deviation = $\sigma$ = 4

Part 1) Proportion of students who scored atleast 25 points

Since, the test scores are normally distributed we can use z scores to find this proportion.

We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:

P(X ≥ 25)

We can convert this value to z score and use z table to find the required proportion.

The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

Using the values, we get:

$z=\frac{25-22}{4}=0.75$

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P(X ≥ 25) is equivalent to P(z ≥ 0.75)

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P(X ≥ 25) = P(z ≥ 0.75), we can conclude:

The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%

Part 2) 31st percentile of the test scores

31st percentile means 31%(0.31) of the students have scores less than this value.

This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.

Using the z table, the z score for 31st percentile comes out to be:

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Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.

Using the values, we get:

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Thus, a test score of 20 represent the 31st percentile of the distribution.

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2 years ago

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

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In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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2 years ago

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Answer:

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