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Nina [5.8K]
3 years ago
15

What is 11.11 as a mixed number ?

Mathematics
2 answers:
zvonat [6]3 years ago
7 0
11.11=11+0.11=11+\frac{11}{100}=\boxed{11 \frac{11}{100}}
maks197457 [2]3 years ago
6 0
11.11 = 1111/100 and simplifies to: 11 11/100
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An auto travels at a rate of 25 km/hr for 4 minutes, then at 50 km/hr for 8 minutes, and finally at 20 km/hr for 2 minutes. find
ankoles [38]

We know that the formula for calculating distance is:

distance d = velocity v * time t

 

Calculating for each interval:

 

1: 25 km/hr for 4 minutes

d1 = (25 km / 60 min) (4 min) = 1.67 km

 

2: 50 km/hr for 8 minutes

d2 = (50 km / 60 min) (8 min) = 6.67 km

 

3: 20 km/hr for 2 minutes

d2 = (20 km / 60 min) (2 min) = 0.67 km

 

The total distance covered is:

d = d1 + d2 + d3

d = 1.67 + 6.67 + 0.67

d = 9 km

 

The average speed is:

s = 9 km (1000 m / km) / (14 min * 60 s / min)

s = 10.72 m / s

3 0
3 years ago
Mark Went to the store and purchase nine pairs of socks and three packages of undershirts for $8.28 each he also purchase a pair
agasfer [191]

Answer:

He paid $2.35 for each pair of socks.

An equation using the variable S is

$77.99 = 9S + (3 × $8.28) + $32

Step-by-step explanation:

Let S be the cost of a pair of socks

From the question, Mark purchase nine pairs of socks, that is, the total cost of the pairs of socks is 9S

He purchase three packages of undershirts for $8.28 each; the total cost of the three packages of shirt will be 3 × $8.28 = $24.84

and he also purchase a pair of pants for $32

Since he spent a total of $77.99, we can write that

$77.99 = 9S + (3 × $8.28) + $32

Now, we can determine S, the cost of a pair of socks

$77.99 = 9S + (3 × $8.28) + $32

$77.99 = 9S + $24.84 + $32

$77.99 = 9S + $56.84

9S = $77.99 - $56.84

9S = $21.15

S = $21.15/9

S = $2.35

Hence, he paid $2.35 for each pair of socks.

7 0
4 years ago
40=9x-5. <br>X=<br><br>liner equation
garik1379 [7]
X=5 hope this worked out for u.
8 0
4 years ago
Read 2 more answers
Which of the following is an example of a rational value between 5 and 10?
sdas [7]

Answer:

vStep-by-step explanation:

8 0
3 years ago
A fair coin is tossed repeatedly with results Y0, Y1, Y2, . . . that are 0 or 1 with probability 1/2 each. For n ≥ 1 let Xn = Yn
Gekata [30.6K]

Answer:

False. See te explanation an counter example below.

Step-by-step explanation:

For this case we need to find:

P(X_{n+1} = | X_n =i, X_{n-1}=i') =P(X_{n+1}=j |X_n =i) for all i,i',j and for X_n in the Markov Chain assumed. If we proof this then we have a Markov Chain

For example if we assume that j=2, i=1, i'=0 then we have this:

P(X_{n+1} = | X_n =i, X_{n-1}=i') =\frac{1}{2}

Because we can only have j=2, i=1, i'=0 if we have this:

Y_{n+1}=1 , Y_n= 1, Y_{n-1}=0, Y_{n-2}=0, from definition given X_n = Y_n + Y_{n-1}

With i=1, i'=0 we have that Y_n =1 , Y_{n-1}=0, Y_{n-2}=0

So based on these conditions Y_{n+1} would be 1 with probability 1/2 from the definition.

If we find a counter example when the probability is not satisfied we can proof that we don't have a Markov Chain.

Let's assume that j=2, i=1, i'=2 for this case in order to satisfy the definition then Y_n =0, Y_{n-1}=1, Y_{n-2}=1

But on this case that means X_{n+1}\neq 2 and on this case the probability P(X_{n+1}=j| X_n =i, X_{n-1}=i')= 0, so we have a counter example and we have that:

P(X_{n+1} =j| X_n =i, X_{n-1}=i') \neq P(X_{n+1} =j | X_n =i) for all i,i', j so then we can conclude that we don't have a Markov chain for this case.

6 0
3 years ago
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