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Nesterboy [21]
3 years ago
12

IF PROBABILITY RANDOMLY CHOSEN ATHLETE IS A SWIMMER IS 0.65, THEN WHAT IS THE PROBABILITY CHOSEN ATHLETE IS NOT A SWIMMER? GIVE

ANSWER AS A DECIMAL
Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

0.35

Step-by-step explanation:

a randomly chosen person's chance of being an athlete is 100%, or 1.00

Therefore: if the probability of a randomly chosen athlete is a swimmer is 0.65 or 65%, The probability of a randomly chosen athlete to NOT be a swimmer is:

1.00 - 0.65 = 0.35 or 35%

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What are the possible values of x in 8x2 + 4x = -1?
Westkost [7]

Answer:

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

Step-by-step explanation:

8x^2+4x+1=0

Using Quadratic formula to solve this equation:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a = 8 \,\, b = 4\,\, c = -1\\Putting \,\, values \,\, in \,\, the\,\, equation\\x= \frac{-4\pm\sqrt{(4)^2-4(8)(-1)}}{2(8)}\\x= \frac{-4\pm\sqrt{16+32}}{16}\\x= \frac{-4\pm\sqrt{48}}{16}\\x= \frac{-4+ \sqrt{48}}{16} \,\, and \,\, x= \frac{-4- \sqrt{48}}{16}\\x= \frac{-1+ \sqrt{3}}{4} \,\, and \,\, x= \frac{-1- \sqrt{3}}{4}

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

4 0
3 years ago
Pls someone find the answer to this question...i have exam tomorrow​
s2008m [1.1K]

Answer: 29.2cm

Step-by-step explanation: perimeter of parallelogram; 2(a + b) where a = base, b = altitude

a = 8cm

b = 6.6cm

Perimeter = 2(8 + 6.6) = 2(14.6)

Perimeter = 29.2cm

8 0
3 years ago
Leo bought a car for x dollars one year later the value of the car was 0.88x.which exspression is another way to describe the ch
Furkat [3]
Well, the car depreciated in value by 12%, so you could say
x-12%, or just
x-0.12.
5 0
3 years ago
(5/6)^4x=(36/25)^9-x, please help solve for x, and the 9-x is all superscript
stiks02 [169]
\frac{36}{25}=\frac{6^2}{5^2}=\left(\frac{6}{5}\right)^2=\left(\frac{5}{6}\right)^{-2}\\\\therefore:\\\\\left(\frac{5}{6}\right)^{4x}=\left[\left(\frac{5}{6}\right)^{-2}\right]^{9-x}\\\\\left(\frac{5}{6}\right)^{4x}=\left(\frac{5}{6}\right)^{-2(9-x)}\iff4x=-2(9-x)\\\\4x=-2\cdot9-2\cdot(-x)\\\\4x=-18+2x\ \ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\2x=-18\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=-9}


Use:\\a^{-n}=\left(\frac{1}{a}\right)^n\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\\left(\frac{a}{b} \right)^n=\frac{a^n}{b^n}
6 0
2 years ago
Read 2 more answers
Did I get it correct?
VikaD [51]

Answer:

No they are proprotional.

Step-by-step explanation:

the tables eqaution is

y=3x+1

7 0
2 years ago
Read 2 more answers
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