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Butoxors [25]
3 years ago
6

PLEASE HELP ASAP ILL GIVE 40 POINTS TO ANYONE THAT WILL ANSWER CORRECTLY JUST LOOK AT SCREENSHOT BELOW

Mathematics
2 answers:
AnnyKZ [126]3 years ago
8 0
129. Just did the math
maks197457 [2]3 years ago
5 0

Answer: 129

Step-by-step explanation:

5^4/5 + 4

= 5^(4-1) + 4

= 5^3 + 4

= 125+4 = 129

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Which shows the equation y – x = 3 written in slope-intercept form? Question 1 options: y – x – 3 = 0 y – 2 = x + 3 y = 3x y = x
Kazeer [188]
Y=x+3
you need to add the x to the opposite side to get y by itself
4 0
3 years ago
In the figure below, XZ = DF and
Sav [38]
Answer is D
XY =  DE
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6 0
4 years ago
Barry is trying to calculate the distance between point E(3, 1) and point F(4, 7). Which of the following expressions will he us
Masteriza [31]
Barry will use the distance formula:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
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7 0
3 years ago
Read 2 more answers
For each student in a certain class, a teacher adjusted the student's test score using the formula y = 0.8x + 20, where x is the
alexandr402 [8]

Answer:

Option B) 16

Step-by-step explanation:

We are given the following in the question:

y = 0.8x + 20

where x is the student's original test score and y is the student's adjusted test score.

We have to find the standard deviation of the adjusted test scores of the students in the class, if the standard deviation of the original test scores of the students in the class was 20.

We know that:

  • Adding a constant to each value in a data set does not change the value of the standard deviation.
  • Multiplying each value in a data set by a constant also multiplies the standard deviation by that constant.

Thus, if we add 20 to each data set then, the standard deviation does not change.

But multiplying each score by 0.8, changes the standard deviation 0.8 times.

Thus, we can write:

\text{Standard deviation of the adjusted test scores}\\ = \text{Standard deviation of the original test scores of the students}\times 0.8\\= 20\times 0.8\\= 16

Thus, standard deviation of adjusted score is 16.

5 0
3 years ago
Determime the intercepts of the line that passes through the following points. (-6,5);(-3,3);(0,1)
butalik [34]

well for a line, to get its slope all we need is two points, so let's use (-6, 5) and (0, 1), and get the equation of it.


\bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{5})\qquad
(\stackrel{x_2}{0}~,~\stackrel{y_2}{1})
\\\\\\
slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-5}{0-(-6)}\implies \cfrac{1-5}{0+6}\implies -\cfrac{5}{6}


\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-5=-\cfrac{5}{6}[x-(-6)]\implies y-5=-\cfrac{5}{6}(x+6)
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\textit{to get the x-intercept, we set y = 0, solve for \underline{x}}
\\\\\\
0-5=-\cfrac{5}{6}(x+6)\implies -30=5x+30\implies -60=5x
\\\\\\
\cfrac{-60}{5}=x\implies -12=x~\hfill \boxed{\stackrel{x-intercept}{(-12,0)}}


now, where's the y-intercept of that line?  well, to get the y-intercept, we set x = 0 and solve for "y"....hmmmm wait a second, notice (0, 1), x = 0, y = 1, that's the y-intercept already.

7 0
4 years ago
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