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2 years ago

I need help pls What is the value of x?

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

8 0

Answer:

Step-by-step explanation:

the value of x, bring the variable to the left side and bring all the remaining values to the right side

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ANTONII [103]

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3 years ago

Write a ratio for the situation in three ways, comparing the first quantity to the second quantity. In Largo City, about one in

saul85 [17]

Answer:

Step-by-step explanation:

1/9

(1*4) / (9*4) = 4/36

(1*4) / (9*4) = 5/45

The first one is the ratio as it is stated in the problem.

The second one multiplies top and bottom by 4

The third one multiplies top and bottom of the top ratio by 5.

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3 years ago

How to solve <br><br> 100x0.25

mart [117]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Can any kind soul help me ASAP!

CaHeK987 [17]

Answer:

Hope it helps u

Step-by-step explanation:

As we know that ,

Mean = sum of the terms/ numbers of terms

But here grouped data is given so , we use the formula

Mean=∑[f. m]/ ∑f

where f is frequency and m is mid point of each height ,

Now first we have to find the mid point of each interval, where

midpoint of each interval = (lower boundary + upper boundary)/2

m1=(150+154)/2 = 152

m2=(155+159)/2= 157,now found other by same formula, for each interval

m3= 162

m4= 167

m5=172 Now we find the midpoint of each interval ,so now

∑[f. m]=f1*m1+f2*m2+f3*m3+f4*m4+f5*m5

now putting the values of each frequency for given interval and midpoint of each interval we will get,

∑[f. m]=456+942+1296+167*x+344 = 167*x+3038

Now find,

∑f=f1+f2+f3+f4+f5

∑f=19+x

Now we have,

∑[f. m]=167*x+3038

∑f=19+x

also given mean height=161.6 cm

putt these values in above equation we get,

161.6= $\frac{167*x+3038}{19+x}$

now solve this ,

161.6(19+x)=167*x+3038

3070.4+161.6*x=167*x+3038

3070.4-3038=167*x-161.6*x

32.4=5.4*x

x=32.4/5.4

3 0

3 years ago

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo

noname [10]

Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}$

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}$

and let's check the reciprocal negative of that,

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6$

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}$

6 0

3 years ago