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Lady_Fox [76]
3 years ago
6

Two triangles with the same corresponding side lengths will be congruent is known as the

Mathematics
2 answers:
MrRissso [65]3 years ago
8 0
.....corresponding sides 

krek1111 [17]3 years ago
4 0
Yes I would go with the same thing.
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PLZ HURRY ITS TIMEDDDDD
GaryK [48]

A. 1/5k - 2/3j and -2/3j + 1/5k

Step-by-step explanation:

Equivalent expressions are expressions that are the same even though they may appear different.When you plug in a value to represent a variable in these expression, they will give same answer when simplified.

In

1/5k - 2/3j and -2/3j + 1/5k you notice the operation signs in the terms has been maintained though the position of the terms shifted.

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Equivalent expressions :brainly.com/question/280220

Keywords: expressions, equivalent

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17. Its answer C i wil tell you the other one in a sec
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What is 38 x 4 = ? <br> i will give away 100 points for next question IF RIGHT
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Answer:

152

Step-by-step explanation:

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The manager of a music store has kept records of
Marysya12 [62]

Answer:

(a) P(x\le 3) = 0.75

(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

(c) P(x \ge 5) = 0.10

(d) P(x = 1\ or\ 2) = 0.55

(e) P(x > 2) = 0.45

Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

Solving (e): Probability of more than 2 CDs

Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

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Circle Q is a translation of circle P, 6 units up and <span>Circle Q is a dilation of circle P with a scale factor of 7 are both correct answers.

Similar figures can result when a figure is translated (moved or slid) to a different position or when it remains the same shape but in a different size (dilation).</span>
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