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seraphim [82]
3 years ago
14

1. How many juniors like to watch track?

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0

Answer:

1. 3

2. 23

3. 50

4. 26%

5. 24%

Step-by-step explanation:

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What is the unit rate and state rate of 253 for 20 hours?
Vilka [71]
The unit rate is 12.65 per hour. 
6 0
4 years ago
Could someone Help me please!
Aliun [14]
C = 2 pi r
C = 2 x 3.14 x 5 
C = 31.4 yd

A = pi r^2
A = 3.14 x 5^2
A = 3.14 x 25
A = 78.5 yd^2
4 0
3 years ago
Find all possible values of the digits Y, E, A, R if YYYY - EEE + AA - R = 1234, and different letters represent different digit
schepotkina [342]

Answer:

Y:  2

E:  9

A:  1

R:  0

Step-by-step explanation:

If we used the value  Y = 2,   E = 9  , A = 1  and R = 0

in the given equation i.e

YYYY - EEE + AA - R = 1234.

From LHS,

Putting the value of Y,E, A and R in the given question we get

2222-999+11-0\\=\ 1223\ + 11\\=\ 1234

=RHS

Therefore\  the\  value\  of \   Y, E, A, R\  is\  2\ , \  9 , \ 1\   and\  0

7 0
4 years ago
The value of n from the set {6, 7, 8, 9} that holds true for 4n − 12 < 2n + 2 is .
IrinaVladis [17]

Answer:

6

Step-by-step explanation:

4n - 12 < 2n + 2 is for n=6

4×6 -12 < 2×6 +2

24 - 12 < 12 + 2

12 < 14 correct

for n=7

4×7 - 12 < 2×7 + 2

16 < 16 incorrect

for n=8

4×8 - 12 < 2×8 + 2

20 < 18 incorrect

for n=9

4×9 - 12 < 2×9 + 2

24 < 20 incorrect

so, only for n=6 is the expression true.

3 0
3 years ago
Assuming boys and girls are equally​ likely, find the probability of a couple having a baby girl when their sixth child is​ born
Ivanshal [37]

Answer:  The required probability of having 6th girl is 0.5.

Step-by-step explanation:  Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is​ born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B}  and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.

Thus, the required probability of having 6th girl is 0.5.

3 0
3 years ago
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