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Pie
1 year ago
15

In a two-digit number, the units digit is 5 more than the tens digit. The number itself is three times the sum of its digits. Wh

at is this number?
Mathematics
1 answer:
Alenkasestr [34]1 year ago
7 0

Answer:

The number is 27.

Step-by-step explanation:

Since the units digit has to be 5 more than the tens digit, there are only four possible two-digit numbers which satisfy this condition. They are 16, 27, 38, and 49.

Next, the digits can be added and multiplied by three to check that the answer is the same two-digit number.

3 (1 + 6) = 21 does not work because 16 isn't the same as 21.

3 (2 + 7) = 27 does work because 3 * 9 is 27.

3 (3 + 8) = 33

3 (4 + 9) = 39

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Suppose that 0.5% of all people have a disease. There is testing available for this disease. If the person being tested has the
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Answer:

98%

Step-by-step explanation:

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3 years ago
He bought four violas for 40$ each. Later he bought one whistle for 100$. After that, he returned one viola. Write the total cha
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40*3=120
100+120=220
Change is £40
6 0
3 years ago
Evaluate the expression. StartFraction 9 factorial Over 3 factorial EndFraction 3 6 60,480 362,874
REY [17]

Answer:

60,480 is the correct answer.

Step-by-step explanation:

First of all, let us have a look at the <em>formula of factorial of a number 'n'</em>:

n! = n \times (n-1) \times (n-2) \times ...... \times 1

i.e. multiply n with (n-1) then by (n-2) upto 1.

<em>Keep on subtracting 1 from the number and keep on multiplying until we reach to 1.</em>

<em></em>

So, 9! can be written as: 9 \times 8 \times 7 \times ...... \times 1

Similarly 3! can be written as: 3 \times 2 \times 1

Re-writing 9 ! :

9 \times 8 \times 7 \times ...... 3 \times 2 \times 1\\\Rightarrow 9 \times 8 \times 7 \times ...... 3 !

Now, the expression to be evaluated:

\dfrac{9!}{3!} = \dfrac{9 \times 8 \times 7 \times ..... \times 3!}{3!}\\\Rightarrow 9 \times 8 \times 7 \times 6 \times 5 \times 4\\\Rightarrow 60480

5 0
3 years ago
Read 2 more answers
A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
AnnZ [28]
  <span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed: 
a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
r = d/2 

Thus: 
a = 2*Pi^2*d/T^2 

What forces cause this acceleration for the passenger, at either top or bottom? 

At top (acceleration is downward): 
Weight (m*g): downward 
Normal force (Ntop): upward 

Thus Newton's 2nd law reads: 
m*g - Ntop = m*a 

At top (acceleration is upward): 
Weight (m*g): downward 
Normal force (Nbottom): upward 

Thus Newton's 2nd law reads: 
Nbottom - m*g = m*a 

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame: 
Ntop = m*(g - a) 
Nbottom = m*(g + a) 


Substitute a: 
Ntop = m*(g - 2*Pi^2*d/T^2) 
Nbottom = m*(g + 2*Pi^2*d/T^2) 

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground): 
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g) 
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g) 

Simplify: 
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2) 
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2) 

Data: 
d:=22 m; T:=12.5 sec; g:=9.8 N/kg; 

Results: 
Ntop/(m*g) = 71.64%...she feels "light" 
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
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3 0
3 years ago
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