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3 years ago

9

Worth a lot of points! please help and explain your answer tysm

1 answer:

Arte-miy333 [17]3 years ago

6 0

Answer:

option c is incorrect

because the value of tan B will be perpendicular/base

so tan B=5/12.

mark me brainlist

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the length of a rectangle is 7 inches more than its width, if the perimiter of the rectangle is 66 inches, find its dimensions.

Maurinko [17]

Perimeter of rectangle=66
length of rectangle=L
width of rectangle=w
P of a rect.= 2(length)+ 2(width)
66= 2L+2w

if the length is 7in more than the width, then
L=7+w

Now we will substitute 7+w in for L. Here is our new equation:

66=2(7+w) + 2w

Solve for w

66=14+2w+2w
66=14+4w
52=4w
w=13
L=7+13, so L=20

I hooe this is explained well enough

3 0

4 years ago

Are the expressions 2(3x-6) and 2(3x) - 6 equivalent? Explain your reasoning.

Sedbober [7]

Answer:

Step-by-step explanation:

No

2(3x-6)=6x-12

2(3x)-6=6x-6

4 0

3 years ago

Read 2 more answers

Bill is making accessories for the soccer team. He uses 611.66 inches of fabric on headbands for 31 players and 3 coaches. He al

Sliva [168]

Answer:

26.8 inches of fabrics was used on headband and wristband for each player

Step-by-step explanation:

Headbands for 31 players and 3 couches = 611.66 inches of fabric

Total number of people = 31 players + 3 couches

= 31 + 3

= 34 people

Fabrics per person's headbands = Total inches of fabrics for headbands / total number of people

= 611.66 inches of fabric / 34 people

= 17.99 inches of fabrics per person

Wristband for 31 players = 273.11 inches of fabric

Fabric per person's wristband = total inches of fabrics for wristband / number of players

= 273.11 inches of fabric / 31

= 8.81 inches per person

How much fabric was used on a headband and wristband for each player?

= Fabrics per person's headbands + Fabric per person's wristband

= 17.99 inches + 8.81 inches

= 26.8 inches of fabrics was used on headband and wristband for each player

4 0

3 years ago

789091 round to the nearest ten thousand

Schach [20]

It is equal to 790000 rounded to the nearest ten thousand

5 0

4 years ago

Read 2 more answers

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing

noname [10]

Answer:

Expectation is equal to 4.75

Variance is equal to 3.02

Step-by-step explanation:

Consider the attached image.

Let $x_i$ denotes each of the outcome.

Expectation $=E(X)$ = ∑ $x_if(x_i)$

$=1(\frac{1}{12}) +2(\frac{1}{12}) +3(\frac{1}{12}) +4(\frac{1}{12}) +5(\frac{1}{12}) +6(\frac{7}{12})$

$=\frac{1}{12}+\frac{2}{12}+\frac{3}{12}+\frac{4}{12}+\frac{5}{12}+\frac{42}{12}\\\\=\frac{57}{12}\\\\=4.75$

$E(X^2)=$ ∑ $x_i^2f(x_i)$

$=1^2(\frac{1}{12}) +2^2(\frac{1}{12}) +3^2(\frac{1}{12}) +4^2(\frac{1}{12}) +5^2(\frac{1}{12}) +6^2(\frac{7}{12})$

$=(\frac{1}{12}) +4(\frac{1}{12}) +9(\frac{1}{12}) +16(\frac{1}{12}) +25(\frac{1}{12}) +36(\frac{7}{12})\\\\=\frac{307}{12}$

Variance $=E(X^2)-[E(X)]^2$

$=(\frac{307}{12})-(\frac{57}{12})^2\\\\ =\frac{307}{12}-\frac{3249}{144}\\\\ =\frac{3684-3249}{144}\\\\ =\frac{435}{144}\\\\ =3.02$

3 0

3 years ago