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gizmo_the_mogwai [7]
3 years ago
5

how is the process of rewriting literal equations similar to the process of solving equations in one variable?

Mathematics
1 answer:
levacccp [35]3 years ago
4 0
Because in both ways, u are isolating a variable by inverse operations.
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Experimental/theoretical
Lera25 [3.4K]

Answer:

A.200

Step-by-step explanation:

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Match the exponential expression with its rule:
mina [271]
1.=d 
2.=a
5.=b
4.=e
3.=c
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If the perimeter is 168 and the width is 6 feet longer than thr length what are the dimensions
Ksenya-84 [330]

\bf \textit{perimeter of a rectangle}\\\\ P=2(L+w)~~ \begin{cases} L=length\\ w=width\\ ----------\\ P=168\\ w=\stackrel{\textit{6 feet longer than L}}{L+6} \end{cases}\implies 168=2(L+\stackrel{w}{L+6}) \\\\\\ 84=2L+6\implies 78=2L\implies \cfrac{78}{2}=L\implies \boxed{39=L} \\\\\\ w=39+6\implies \boxed{w=45}

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3 years ago
Find the measures of the legs of isosceles triangle ABC if AB= 2x+4,BC= 3x-1,AC=x+1 and the perimeter of the triangle ABC is 34
Naddika [18.5K]
(2x+4) +(3x-1) +(x+1)=34
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5 0
3 years ago
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The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged
Temka [501]

Answer:

a) 0.283  or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z=\frac{X-M}{s} where

  • X is the left atrial diameter value we are looking for its z-score
  • M is the mean left atrial diameter of healthy children (26.7 mm)
  • s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*=\frac{24-26.7}{4.7} ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*=\frac{32-26.7}{4.7} ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)=\frac{30-26.7}{4.7} ≈ 0.702

z(25)=\frac{25-26.7}{4.7} ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84=\frac{X-26.7}{4.7}  solving this equation for X we get X=30.648

5 0
3 years ago
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