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2 years ago

14

a computer artist created a rectangular image measuring 55 125 scale factor of 2.5 what are the dimensions of the dilated image?

quick!!

1 answer:

lesya692 [45]2 years ago

4 0

Answer:

sex

Step-by-step explanation:

69

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The formula for the volume sphere is V=4/3pie r^3 what is the formula solved for r?

oksian1 [2.3K]

Answer:

$r = \sqrt[3]{\frac{3V}{4 \pi}}$

Step-by-step explanation:

From the formula of volume of a sphere we have to isolate "r" on one side of the equation i.e. we have to make "r" the subject of the equation.

$V=\frac{4}{3} \pi r^{3}\\\\ \text{Multiplying both sides by 3/4 we get}\\\\\frac{3V}{4} = \pi r^{3}\\\\ \text{Dividing both sides by } \pi \\\\ \frac{3V}{4 \pi} = r^{3}\\\\\text{Takeing cube root of both sides}\\\\\sqrt[3]{\frac{3V}{4 \pi}} = r$

Therefore:

$r = \sqrt[3]{\frac{3V}{4 \pi}}$

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4 years ago

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1 4 6 60 7 9 10TIME REMAIN41:42For a snack, Sameer ate a plum with 32 calories and several strawberries with 4 calories each. If

Alex777 [14]

Well, if Sameer's plum was 32 calories, and we know his total snack was 56 calories, all we have to do is:

56-32=24

24 / 4 = 6

Sameer ate one plum (32Kcal) and 6 strawberries (4Kcal each)

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3 years ago

What is the point 2,1 dilated by the scale factor of 2.5

Inga [223]

The answer would be -0.4

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3 years ago

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A car drives 195 miles in 3 hours and 15 mins.<br> what is the average speed of the car?

maria [59]

Answer:

one mile per hour

Step-by-step explanation:

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3 years ago

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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:

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

$P(x > 2) = 0.45$

3 0

3 years ago