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3 years ago

15

When representing a frequency distribution with a bar chart, which of these

1 answer:

gogolik [260]3 years ago

7 0

The answer would be C

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PLZ HELP!!!!!!! (20 points)

AlekseyPX

Answer:

Neither A nor B

Step-by-step explanation:

The set of pairs, where the front member of every sequence pair must be unique.

3 0

3 years ago

Nora's paper airplane flew 9 1/6 feet. That was 2 3/4 feet farther than Max's plane flew. Write and solve an equation to show ho

stepan [7]

Answer:

Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

Step-by-step explanation:

As Nora's paper airplane flew 9 1/6 feet

i.e.

$9\frac{1}{6}=\frac{55}{6}$

Nora's paper was farther than Max's plane flew by = 2 3/4 feet

i.e.

$2\frac{3}{4}=\frac{11}{4}$

Thus the equation to show how far Max's paper airplane flew is:

$\:\:9\frac{1}{6}-2\frac{3}{4}$

$=\frac{55}{6}-\frac{11}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}6,\:4:\quad 12$

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}$

$=\frac{110}{12}-\frac{33}{12}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{110-33}{12}$

$\mathrm{Subtract\:the\:numbers:}\:110-33=77$

$=\frac{77}{12}$

$=6\frac{5}{12}$

Therefore, Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

6 0

3 years ago

Help pls FAST !! Plsss

Lorico [155]

Answer:

$\sqrt{17}$ , 4.23 , 9/2

Step-by-step explanation:

$\sqrt{17} = 4.1231\\$

9/2 = 4.5

4.23

7 0

3 years ago

What is the mixed number of 96/8

lions [1.4K]

The answer is 12 because i divide 96 / 6

(sorry i dont have division sign)

3 0

3 years ago

Please solve these questions for me. i am having a difficult time understanding.

s2008m [1.1K]

Answer:

1) AD=BC(corresponding parts of congruent triangles)

2)The value of x and y are 65 ° and 77.5° respectively

Step-by-step explanation:

1)

Given : AD||BC

AC bisects BD

So, AE=EC and BE=ED

We need to prove AD = BC

In ΔAED and ΔBEC

AE=EC (Given)

$\angle AED = \angel BEC$ ( Vertically opposite angles)

BE=ED (Given)

So, ΔAED ≅ ΔBEC (By SAS)

So, AD=BC(corresponding parts of congruent triangles)

Hence Proved

2)

Refer the attached figure

$\angle ABC = 90^{\circ}$

In ΔDBC

BC=DC (Given)

So, $\angle CDB=\angle DBC$ (Opposite angles of equal sides are equal)

So, $\angle CDB=\angle DBC=x$

So, $\angle CDB+\angle DBC+\angle BCD = 180^{\circ}$ (Angle sum property)

x+x+50=180

2x+50=180

2x=130

x=65

So, $\angle CDB=\angle DBC=x = 65^{\circ}$

Now,

$\angle ABC = 90^{\circ}\\\angle ABC=\angle ABD+\angle DBC=\angle ABD+x=90$

So, $\angle ABD=90-x=90-65=25^{\circ}$

In ΔABD

AB = BD (Given)

So, $\angle BAD=\angle BDA$ (Opposite angles of equal sides are equal)

So, $\angle BAD=\angle BDA=y$

So, $\angle BAD+\angle BDA+\angle ABD = 180^{\circ}$ (Angle Sum property)

y+y+25=180

2y=180-25

2y=155

y=77.5

So, The value of x and y are 65 ° and 77.5° respectively

8 0

3 years ago