Answer:
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given total number of students n(T) = 150
Given 125 of them are fluent in Swahili
Let 'S' be the event of fluent in Swahili language
n(S) = 125
The probability that the fluent in Swahili language
Let 'E' be the event of fluent in English language
n(E) = 135
The probability that the fluent in English language
n(E∩S) = 95
The probability that the fluent in English and Swahili
<u><em>Step(ii):</em></u>-
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = P(S) + P(E) - P(S∩E)
= 0.833+0.9-0.633
= 1.1
<u><em>Final answer:-</em></u>
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Answer:
C 9i
D -9i
Step-by-step explanation:
sqrt(-81)
sqrt(81) sqrt(-1)
we know that sqrt(-1) = i
±9i
Answer:
the answer is 12
Step-by-step explanation:
Simplify both sides of the equation.
12(b−10)+2b=25
(12)(b)+(12)(−10)+2b=25(Distribute)
12b+−5+2b=25
(12b+2b)+(−5)=25 (Combine Like Terms)
52b+−5=25
52b−5=25
add 5 to both sides.
52b−5+5=25+5
52b=30
multiply both sides by 2/5.
(25)*(52b)=(25)*(30)
b=12
Answer:
Step-by-step explanation:
![x_{1}=7 \ ; y_{1} = -4\\\\x_{2} = -8 \ ; y_{2}=5\\\\CD = \sqrt{(-8-7)^{2}+(5-[-4])^{2}}\\\\=\sqrt{(-15)^{2}+(5+4)^{2}}\\\\=\sqrt{225+(9)^{2}}\\\\=\sqrt{225+81}\\\\=\sqrt{306}\\\\=\sqrt{3*3*2*17}\\\\=3\sqrt{34}](https://tex.z-dn.net/?f=x_%7B1%7D%3D7%20%5C%20%3B%20y_%7B1%7D%20%3D%20-4%5C%5C%5C%5Cx_%7B2%7D%20%3D%20-8%20%5C%20%3B%20y_%7B2%7D%3D5%5C%5C%5C%5CCD%20%3D%20%5Csqrt%7B%28-8-7%29%5E%7B2%7D%2B%285-%5B-4%5D%29%5E%7B2%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%28-15%29%5E%7B2%7D%2B%285%2B4%29%5E%7B2%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B225%2B%289%29%5E%7B2%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B225%2B81%7D%5C%5C%5C%5C%3D%5Csqrt%7B306%7D%5C%5C%5C%5C%3D%5Csqrt%7B3%2A3%2A2%2A17%7D%5C%5C%5C%5C%3D3%5Csqrt%7B34%7D)
Distance = 
if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.
so, any when we're referring to even roots like ![\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B2%5D%7B~~%7D%2C%5Csqrt%5B4%5D%7B~~%7D%2C%5Csqrt%5B6%5D%7B~~%7D....)
, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.
now, that's is not true for odd roots like ![\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B~~%7D%2C%5Csqrt%5B5%5D%7B~~%7D%2C%5Csqrt%5B7%5D%7B~~%7D....)
, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.
![\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B27%7D%5Cimplies%20%5Csqrt%5B3%5D%7B3%5E3%7D%5Cimplies%203%5Cqquad%20because%5Cqquad%20%283%29%283%29%283%29%3D27%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%28-3%29%28-3%29%28-3%29%5Cne%2027~%5Chspace%7B8em%7D%28-3%29%28-3%29%28-3%29%3D-27%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B-125%7D%5Cimplies%20%5Csqrt%5B3%5D%7B-5%5E3%7D%5Cimplies%20-5%5Cqquad%20because%5Cqquad%20%28-5%29%28-5%29%28-5%29%3D-125%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%285%29%285%29%285%29%5Cne%20-125~%5Chspace%7B10em%7D%285%29%285%29%285%29%3D125)
so, when the root is an odd root, you will always get only one number that will produce the radicand.
