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julsineya [31]
3 years ago
12

Daisy went to an electronics store and paid $78.99 per GPS navigator for five GPS navigators and a flat rate fee for installatio

n for a total of $428.45. Rebecca went to another electronics store and paid $85.49 per GPS navigator for five GPS navigators plus a per GPS navigator installation fee for a total of $464.35. What is the difference in the amounts they paid in installation charges?
A. $3.40
B. $4.40
C. $2.40
D. $5.40
Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0
Daisy:
78.99 x 5 = 394.95
428.45 - 394.95 = 33.50 installation fee

Rebecca:
85.49 x 5 = 427.45
464.35 - 427.45 = 36.90 installation fee

difference : 36.90 - 33.50 = 3.40

Answer is A
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Consider the series 7/9+7/27+7/81+7/243
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Answer:

=1 37/243

Step-by-step explanation:

the series is that the numerator is all 7

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Through (5,5) parallel to 5x-5
Crazy boy [7]

Answer:

y=5x-20

Step-by-step explanation:

For the equation y=5x-5, the slope is 5 and the slope of the new line will be 5 because parallel lines have the same slope. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.

(y-5)=5(x-5)

y-5=5x-25

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POSSIBLE OUTCOMES!!!
Ganezh [65]

The probability of drawing two blue marbles if the first marble is not replaced is 1/5

<h3>How to determine the probabilities?</h3>

<u>The probability of tossing a head and drawing a red marble</u>

The given parameters are:

White = 1

Blue =3

Red = 2

Total = 6

The probability of a head is

P(Head)= 1/2

The probability of drawing a red marble is

P(Red)= 2/6 = 1/3

The required probability is

P = P(Head) * P(Red)

This gives

P = 1/2 * 1/3

P =1/6

<u>The probability of drawing two blue marbles if the first marble is not replaced.</u>

Here, we have:

P(B1) = 3/6 = 1/2

P(B2) = 2/5

The required probability is

P = P(B1) * P(B2)

This gives

P = 1/2 * 2/5

P =1/5

Hence, the probability of drawing two blue marbles if the first marble is not replaced is 1/5

Read more about probability at:

brainly.com/question/24756209

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5 0
2 years ago
Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0
3 years ago
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