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mote1985 [20]
3 years ago
13

What is the product of 0.6 and 7

Mathematics
2 answers:
Lunna [17]3 years ago
8 0
Product means multiplication.

0.6 * 7 = 4.2
yuradex [85]3 years ago
3 0
0,6·7=6/10·7=42/10=4,2

<em>Answer: 4,2</em>
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Which of the following is not true about the standard error of a statistic?
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Answer:

Only option d is not true

Step-by-step explanation:

Given are four statements about standard errors and we have to find which is not true.

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-- True because population parameter is mean and the statistic are the items.  Hence the differences average would be std error.

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3 years ago
Match each statement (term) with its value in relation to 0 (definition). Match Term Definition A building is 38 feet tall. A) –
Kitty [74]
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7 0
2 years ago
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Help me with this please
Klio2033 [76]

Answer:

No, she's wrong

x\leq\frac{2}{5}

Step-by-step explanation:

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3x+3\leq -2x+5

Subtract the 3 from both side and it will look like...

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4 0
2 years ago
2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?
7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

The solutions to the quadratic equation 2x^2\:-\:5x\:+\:67\:=\:0 are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Step-by-step explanation:

Considering the equation

2x^2\:-\:5x\:+\:67\:=\:0

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

so

\frac{2x^2-5x}{2}=\frac{-67}{2}

x^2-\frac{5x}{2}=-\frac{67}{2}

Lets now solve the equation by completeing the remaining steps

Write equation in the form: x^2+2ax+a^2=\left(x+a\right)^2

Solving for a,

2ax=-\frac{5}{2}x

a=-\frac{5}{4}

\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}

Completing the square

\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

For f^2\left(x\right)=a the solution are f\left(x\right)=\sqrt{a},\:-\sqrt{a}

Solving

x-\frac{5}{4}=\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}

x-\frac{5}{4}=i\sqrt{\frac{511}{16}}       ∵ Applying imaginary number rule \sqrt{-1}=i

x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}

-\frac{5}{4}=i\frac{\sqrt{511}}{4}

x=\frac{5}{4}+i\frac{\sqrt{511}}{4}

Similarly, solving

x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}    ∵ Applying imaginary number rule  \sqrt{-1}=i

x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Therefore, the solutions to the quadratic equation are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

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