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Rainbow [258]
4 years ago
14

Trevor Linden, a former player for the Vancouver Canucks, is 1.93 m tall. On a hockey card, he is 5.4 cm tall. What scale was us

ed to print the hockey card?
Mathematics
1 answer:
Anastaziya [24]4 years ago
7 0

Answer:

scale  used to print the hockey card is 27/965

scale  used to print the hockey card in decimal  is 1/35.74

Step-by-step explanation:

we know that in scale problems

scale is used in drawing to enlarge or reduce size of object by a certain amount

scale = size of object in drawing/ actual size of object

Height of trevor = 1.93 m

we know that  1 m = 100 cm

Height of trevor in cm = 1.93 m*100 = 193 cm (actual size)

(this is done so that unit of measurement is same for card and trevor's height i.e cm)

Trevor's card on hockey card = 5.4 cm(size in drawing)

scale used = Trevor's card on hockey card /Height of trevor in cm

                    = 5.4/193 = 54/ 1930 = 27/965

scale  used to print the hockey card is 27/965

scale  used to print the hockey card in decimal  is 1/35.74

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The equation y 16t 2 18t 405 describes the height in feet of a ball thrown downward at 18 feet per second from a height of 405 f
kumpel [21]

Answer:

Ball hits the ground after 4.5 sec

Step-by-step explanation:

Let a -1, so that the leading coefficient is positive

So our quadratic is

-1 \times (16t^2 +18t -405)

The key coefficients of two binomial variables can be 1 and 16, or 2 and 8, or 4 and 4, for the leading coefficient of 16.

Yet they can't actually be 4 and 4 because the linear (x) term coefficient has to be a multiple of 4, which it isn't and leading coefficients 1 and 16 on the binomial factors is not likely.

So, 2 and 8 taken as the leading coefficients of  two binomial factors.

For constant  405, possible factorizations are (81\times 5, 27 \times 15, 9 \times 45).

(16t^2 +18t -405) = (8t + 45) (2t - 9)

Taking first factor, thus we find  negative value for given time t. But  second time equivalent to zero gives the value of 4.5 for t

Thus ball hits the ground after 4.5 sec

.

3 0
4 years ago
9 and 10 pls help :)
tatuchka [14]

Answer:

Question 9

Option A

Question 10

Option A

<h2>Please mark as brainliest for further answers :)</h2>

5 0
3 years ago
Write an equation that models the sequence 400, 200, 100, 50, ...
Serhud [2]
The answer is d because the amplitude is 400, and it is decreasing by a factor of 1/2.
8 0
3 years ago
Explain why P(A|D) and P(D|A) from the table below are not equal. A 4-column table has 3 rows. The first column has entries A, B
dangina [55]

Answer:

P(A|D) and P(D|A) from the table above are not equal because P(A|D) = \frac{2}{10} and P(D|A) = \frac{2}{8}

Step-by-step explanation:

Conditional probability is the probability of one event occurring  with some relationship to one or more other events

.

P(A|D) is called the "Conditional Probability" of A given D

P(D|A) is called the "Conditional Probability" of D given A

The formula for conditional probability of P(A|D) = P(D∩A)/P(D)

The formula for conditional probability of P(D|A) = P(A∩D)/P(A)

The table

               ↓         ↓       ↓

            :  C     :   D    : Total

→ A      :  6     :    2    :   8

→ B      :  1      :    8    :   9

→Total :  7     :    10  :  17

∵ P(A|D) = P(D∩A)/P(D)

∵ P(D∩A) = 2 ⇒ the common of D and A

- P(D) means total of column D

∵ P(D) = 10

∴ P(A|D) = \frac{2}{10}

∵ P(D|A) = P(A∩D)/P(A)

∵ P(A∩D) = 2 ⇒ the common of A and D

- P(A) means total of row A

∵ P(A) = 8

∴ P(D|A) = \frac{2}{8}

∵ P(A|D) = \frac{2}{10}

∵ P(D|A) = \frac{2}{8}

∵  \frac{2}{10} ≠  \frac{2}{8}

∴ P(A|D) and P(D|A) from the table above are not equal

5 0
4 years ago
Read 2 more answers
Graph the function y=<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20-2-5" id="TexFormula1" title="\sqrt{x} -2-5" alt="\sqrt{
Rashid [163]

Answer:

(3, -4)

Step-by-step explanation:

The correct equation is:

y=\sqrt{x-2}-5

From the list of given points, we have to identify which point lies on the graph. This can be done by using the value of x-coordinates from the given points and see if they result in the corresponding y-coordinate:

For point (1, 4)

y=\sqrt{1-2}-5=\sqrt{-1}-5, This is not equal to 4, so it does not lie on the graph of given function.

For point (3, -4)

y=\sqrt{3-2}-5=\sqrt{1}-5=-4, The answer ans corresponding y-coordinate are the same. This means, (3, -4) point lies on the graph of given function.

Likewise, checking for 3rd and 4th point we find out that they do not lie on the graph.

So the correct answer is (3, -4)

5 0
3 years ago
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