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3 years ago

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A multiple-choice test consists of 10 questions. Each question has answer choices of a ,b c, d ,e , and , and only one of the ch

oices is correct. If a student randomly guesses on each question, what is the probability that he gets at least 2 of them correct?

1 answer:

aalyn [17]3 years ago

4 0

Its easier to find the complement of that instead
so you find the probability of 1 and 0
The .2 in the answer is the probability of each question 1/5
(10 C 0) x .2^0 (1-.2)^10-0=.1074
(10 C 1) x .2^1 (1-.2)^10-1=.2684
.2684+.1074=.3758 and that's the complement
so 1 - .3758= .6242
I too have a stats exam on this tomorrow lol

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aalyn [17]

Answer is
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when x = 2, y = -2(2) + 3 = -4 + 3 = -1

solutions matched the table above

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3 years ago

The value of m, if 3,18,m,42 are in proportion

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If its a proportion then

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3 years ago

Through (-6,4), slope:2

madreJ [45]

Answer:

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3 years ago

Plz help quick How do you do this problem

astraxan [27]

Answer: $\frac{pr}{675}$

Step-by-step explanation:

cant have a negative exponent
$\frac{1}{(3p^{-1} r)^3(5pr^{-2} )^2} \\$
split up (3p^-1 r)^3
rewrite that part of the equation
$\frac{1}{(\frac{3}{p} )^{3} r^{3} }$
raise 3 to the power of 3 = 27
p to the power of 3 = $p^{3}$
Next part of the equation
$(5p)^{2}$
expand that
$5^{2} p^{2}$
simplfy
$25p^{2}$
Next part of the equation
$(r^{-2})^{2}$
multiply the exponents
$r^{-4}$
rewrite the equation
$\frac{1}{\frac{27}{p^{3} }r^{3} (25p^{2})\frac{1}{r^{4} } }$
combine $\frac{27}{p^{3} } , and , r^{3}$ = $\frac{27r^{3} }{p^{3} }$
combine 25 and $\frac{27r^{3} }{p^{3} }$
Multiply
$\frac{675r^{3} }{p^{3} }$
combine $\frac{675r^{3} }{p^{3} }$ and $p^{2}$
$\frac{675r^{3}p^{2} }{p^{3} }$
multiply $\frac{675r^{3}p^{2} }{p^{3} }$ and $\frac{1}{r^{4} }$
$\frac{1}{\frac{675r^{3}p^{2} }{p^{3}r^{4} }}$
cancel the common factors and factor everything out
$r^{3}$ comes out of ${675r^{3}p^{2} }$
$r^{3}$ can also come out of $p^{3} r^{4}$
Rewrite as $\frac{1}{\frac{675p^{2} }{p^{3} r} }$
cancel the common factor of $p^{2}$ and $p^{3}$
factor $p^{2}$ out of $675p^{2}$ by mulitiplying
factor $p^{2}$ out of ${p^{3} r} }$
cancel the common factor
rewrite the expression
$\frac{1}{\frac{675}{pr} }$
and then multiply the numerator and denominator
and you get the answer

7 0

2 years ago