Answer:
Lets take the hypothesis
<em>zα/2 </em>
<em>- zα/2 </em>
Step-by-step explanation:
<em>α/2 = 0.10/2 = 0.05
</em>
<em>z(0.05) = 3.92
</em>
<em>-z(0.05) = -3.92
</em>
9.1/3575
use long division like this
_<u>0.</u>_________
3575|91.000000
we find how many of 3575 fit into the 91
it's too big so we go farther
how many 3575 go into 910
too big so we go farther
how many 3575 go into 9100
the answer is 2 so put that in the correct place and mulitply that and put that in the correct place. then we subtract
_<u>0.</u><u>02</u>_________
3575|91.000000
-<u>71.50</u>
19.50
bring down the next number
find how many go into 19.500
the answe ris 5
_<u>0.</u><u>02</u><u>5</u>_________
3575|91.000000
-<u>71.50</u>
19.500
-<u>17.875</u>
1.625
bring down he next zero ( I fast forward and skip steps for convinience)
__
_<u>0.</u><u>02</u><u>5</u><u>455</u>_________
3575|91.000000
-<u>71.50</u>
19.500
-<u>17.875</u>
1.6250
-<u>1.4300</u>
.19500
-<u>.17875
</u> 16250
-14300
and so on untill infinity so the answe ris 0.0254555555555555 (enless fives)
Answer:2.5×10^-5. (And i think u meant in Scientific Notation)
Step-by-step explanation//Give thanks(and or Brainliest) if helpful (≧▽≦)//
The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial
a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.
b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.
Answer:
x=10
Step-by-step explanation:
2x-7=13
13+7=20
20/2=10
lets see if it works:
2(10)-7-13
20-7=13
it works! which means is 10
please mark me as brainliest!!<3
