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Lyrx [107]
3 years ago
10

Solve for x in the equation x2-5x+1=3

Mathematics
1 answer:
LenKa [72]3 years ago
6 0
x^{2} -5x+1=3

The first thing you should do to solve is set the equation equal to zero.  Subtract 3 from each side:

x^{2} -5x+1=3 \\ \\ x^{2} -5x-2=0

The resulting equation doesn't seem to be factorable.  To solve for x, use the quadratic formula:

x= \frac{-b \pm \sqrt{b^2-4ac} }{2a} \\ \\ x= \frac{-(-5) \pm \sqrt{(-5)^2-4(1 )(-2)} }{2(1)} \\ \\ x= \frac{5 \pm \sqrt{25-4(-2)} }{2} \\ \\ x= \frac{5 \pm \sqrt{25+8} }{2} \\ \\ x= \frac{5 \pm \sqrt{33} }{2} \\ \\ x \approx \frac{5 \pm 5.74}{2} \\ \\ \\ \\ x \approx \frac{5+5.74}{2} \\ \\ x \approx \frac{10.74}{2} \\ \\ x \approx 5.37 \\ \\ \\ \\ x \approx \frac{5-5.74}{2} \\ \\ x \approx \frac{-0.74}{2} \\ \\ x \approx -0.37

To the nearest hundredths place, the solutions to the original equation are x = 5.37 and x = -0.37.
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Time traveler can take 5 books with him and he has 79 books to choose from how many different ways can the books be selected
AveGali [126]

<u>We are given:</u>

The time-traveller has 79 books, but he can only take 5 books

since we need to find the number of ways he can choose 5 books, we know that the order in which he takes the books does NOT matter

Hence, we will use combination

<u>Finding the number of ways:</u>

Since we will use combination, ₇₉C₅

We know that the formula for combination is  (n!) / (r!(n-r)!)

₇₉C₅  = 79! / 5!(74!)

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Therefore, the time traveller can choose 5 books in 22,537,515 different ways

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A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand
Tpy6a [65]

Answer:

We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

<em>We have to test the claim that the population mean is 10.5.</em>

Let, NULL HYPOTHESIS, H_0 : \mu = 10.5  {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, H_a : \mu \neq 10.5  {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

           T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean = 11

            s = sample standard deviation = 2

            \mu = population mean

            n = sample of values = 16

So, <u>test statistics</u> =  \frac{11-10.5}{\frac{2}{\sqrt{16} } } ~ t_1_5

                            = 1

<em>Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.</em>

Therefore, we conclude that the population mean is different from 10.5.

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4 years ago
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