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4 years ago

Determine the given point is a solution to the equation. Show all work. 1) 3x-7y=4; (1,-1)

Mathematics

1 answer:

marusya05 [52]4 years ago

6 0

Answer:

NOT a solution

Step-by-step explanation:

Substitute the values of x and y and see if the equation is satisfied:

3(1) -7(-1) = 3 +7 = 10 ≠ 4 . . . . . . the given point is not on the line

(1, -1) is not a solution to the equation.

_____

(-1, 1) is a solution

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a population has a mean of 25 median of 24 and mode of 26 the standard deviation is 5 the value for the

Arada [10]

Complete Question

A population has a mean of 25, a median of 24, and a mode of 26. The standard deviation is 5. The value of the 16th percentile is _______. The range for the middle 3 standard deviation is _______.

Answer:

The value of the 16th percentile is $x = 20$

The range for the middle 3 standard deviation is $10 \to 40$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 25$

The median is $m = 24$

The mode is $n = 26$

The standard deviation is $\sigma = 5$

Generally the 16th percentile is mathematically represented as

$P(x)= P(\frac{X - \mu }{ \sigma } \le \frac{x- 25 }{5} ) = 0.16[/teGenerally [tex]\frac{X - \mu}{\sigma } = Z(The \ standardized \ value \ of \ X )$

$P(x) = P(Z \le \frac{ x - 25}{5} ) = 0.16$

Now from the normal distribution table the z-score of 0.16 is

z = -1

$P(x) = P(Z \le \frac{ x - 25}{5} ) = P(Z \le -1 )$

=> $\frac{x- 25}{5} = -1$

=> $x- 25 = -5$

=> $x = 25-5$

=> $x = 20$

The range for the middle 3 standard deviation is mathematically represented

$\mu - 3\sigma \to \mu + 3\sigma$

$25 - 3(5 ) \to 25 + 3(5 )$

$10 \to 40$

3 0

3 years ago

Can you help me with #1?

steposvetlana [31]

The answer would be 27. If ABC is 3 times the length of AB then it would be the same for AC.

5 0

3 years ago

Add a explanation with your answer

Alenkasestr [34]

I hope this helps !!

8 0

3 years ago

I need help I’m not sure

Daniel [21]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{-6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-6+3}{2}~~,~~\cfrac{-6+5}{2} \right)\implies \left(\cfrac{-3}{2}~~,~~\cfrac{-1}{2} \right)\implies \left( -1\frac{1}{2}~,~-\frac{1}{2} \right)$