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3 years ago

The solutions of x^2= 16x-28

Mathematics

2 answers:

-BARSIC- [3]3 years ago

6 0

The answer it (4,14)

STALIN [3.7K]3 years ago

5 0

Answer:

Simplifying

x2 = 16x + -28

Reorder the terms:

x2 = -28 + 16x

Solving

x2 = -28 + 16x

Solving for variable 'x'.

Reorder the terms:

28 + -16x + x2 = -28 + 16x + 28 + -16x

Reorder the terms:

28 + -16x + x2 = -28 + 28 + 16x + -16x

Combine like terms: -28 + 28 = 0

28 + -16x + x2 = 0 + 16x + -16x

28 + -16x + x2 = 16x + -16x

Combine like terms: 16x + -16x = 0

28 + -16x + x2 = 0

Factor a trinomial.

(2 + -1x)(14 + -1x) = 0

Subproblem 1

Set the factor '(2 + -1x)' equal to zero and attempt to solve:

Simplifying

2 + -1x = 0

Solving

2 + -1x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-2' to each side of the equation.

2 + -2 + -1x = 0 + -2

Combine like terms: 2 + -2 = 0

0 + -1x = 0 + -2

-1x = 0 + -2

Combine like terms: 0 + -2 = -2

-1x = -2

Divide each side by '-1'.

x = 2

Simplifying

x = 2

Subproblem 2

Set the factor '(14 + -1x)' equal to zero and attempt to solve:

Simplifying

14 + -1x = 0

Solving

14 + -1x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-14' to each side of the equation.

14 + -14 + -1x = 0 + -14

Combine like terms: 14 + -14 = 0

0 + -1x = 0 + -14

-1x = 0 + -14

Combine like terms: 0 + -14 = -14

-1x = -14

Divide each side by '-1'.

x = 14

Simplifying

x = 14

Solution

x = {2, 14}

Step-by-step explanation:

I type fast lol

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A test of H0: p = 0.5 versus Ha: p > 0.5 has the test statistic z = 1.15. Part A: What conclusion can you draw at the 5% sign

Nadya [2.5K]

Answer:

Part A: The null hypothesis failed to be rejected.

Part B: The null hypothesis failed to be rejected.

Step-by-step explanation:

We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.

Part A: If the significance level is 0.05, the conclusion depends on the P-value.

If the P-value is below 5%, the null hypothesis is rejected.

The P-value for this right-tailed tes and z=1.15 is:

$P-value=P(z>1.15)=0.125$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.

This changes the way we calculate the P-value, as we need to compute the two tails.

The P-value is:

$P-value=2P(z>1.15)=0.25$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

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3 years ago

At South High School, 37% of the teachers are between 22 and 30 years old, 25% are between 31 and 40

Nookie1986 [14]

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3 years ago

Find the indicated values g(t)=t^2-t and f(x)=1+x g(f(2)+3)

Semmy [17]

Answer:

g( f(2)+3) = 30

Step-by-step explanation:

step(i):-

Given function g(t)=t²-t and f(x)=1+x

given f(x)=1+x

f(2) = 1 + 2 =3

step(ii):-

g( f(2)+3) = g(3 + 3)

= g(6)

= (6)² - 6

= 36 -6

= 30

Conclusion:-

g( f(2)+3) = 30

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3 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

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