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3 years ago

I cannot understand this help me and I do not know what a scale factor is

Mathematics

2 answers:

kipiarov [429]3 years ago

5 0

Answer:

The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure. Example: Here, XYUV=123=4 . So, the scale factor is 4

Step-by-step explanation:

Olegator [25]3 years ago

4 0

Answer:

the scale factor would be 4

Step-by-step explanation:

also

A scale factor is a number which scales, or multiplies, some quantity. In the equation y = Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a scale factor of two for distance, while cutting a cake in half results in pieces with a scale factor for volume of one half. The basic equation for it is image over preimage.

branliest please also be sure to comment and i'll help with everything again the scale factor of the attachment given is 4

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Find the common ratio for the following sequence. 64, 16, 4, 1, ...

insens350 [35]

Common ratio = a2 / a1 or a3/a2
= 16/64 = 1/4

In short, Your Answer would be 1/4

Hope this helps!

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What other multiplication facts can be used to find the product of 5×6?

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5*6 can be expressed as 5+5+5+5+5+5

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This year, the Start-of-Term feast had three desserts: chocolate eclairs, raspberry tarts, and jam doughnuts. On the dessert tab

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Let
x = chocolate eclairs
y = raspberry tarts
z = jam donuts
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z = (2/3) x
x = (6/5) y
x + y + z = 1320
Solving for y:
z = (2/3) (6/5) y
z = (4/5) y
then
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3 years ago

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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = 0.0083.

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kenny6666 [7]

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