When simplified completely, the product of a monomial and a binomial is <br> a trinomial.

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

Yves bought 420 tropical fish for a museum display. He bought 6 times as many parrot fish as angelfish. How many of each type of

Anna11 [10]

6a=p

6(60)=360

60+360=420

the answer is 60 angelfish and 360 parrot fish.

5 0

4 years ago

Read 2 more answers

What expression is equivalent to 3(x+4)^2

eimsori [14]

Hello there,

Well we are given with 3(x + 4)² and we need to simplify this expression...

write out the expression: 3(x + 4)(x + 4)
then we distribute the parentese: 3(x² +4x +4x +16)
combine like terms: 3(x² + 8x + 16)
distribute the 3 in the parentese: 3x² + 24x + 48

So another way to express this would be 3x² + 24x + 48.

Hope I helped,

Amna

8 0

3 years ago

stephanie spends 3/4 of an hour on homework five nights a week. how many hours does stephanie spend on home work throught out th

AnnyKZ [126]

Answer:

5

Step-by-step explanation:

trtreeewe

7 0

4 years ago

Simon has more money than Kande. if Simon gave Kande K20, they would have the same amount. While if Kande gave Simon $22, Simon

motikmotik

Given parameters;

Let us solve this problem step by step;

Let us represent Simon's money by S

Kande's money by K

Simon has more money than Kande

S > K

if Simon gave Kande K20, they would have the same amount;

if Simon gives $20, his money will be S - 20 lesser;

When Kande receives $20, his money will increase to K + 20

S - 20 = K + 20 ------ (i)

While if Kande gave Simon $22, Simon would then have twice as much as Kande;

if Kande gave Simon $22, his money will be K - 22

Simon's money, S + 22;

S + 22 = 2(K - 22) ------ (ii)

Now we have set up two equations, let us solve;

S - 20 = K + 20 ---- i

S + 22 = 2(K - 22) ; S + 22 = 2K - 44 ---- ii

So, S - 20 = K + 20

S + 22 = 2K - 44

subtract both equations;

-20 - 22 = (k -2k) + 64

-42 = -k + 64

k = 106

Using equation i, let us find S;

S - 20 = K + 20

S - 20 = 106 + 20

S = 106 + 20 + 20 = 146

Therefore, Kande has $106 and Simon has $146

3 0

3 years ago

When simplified completely, the product of a monomial and a binomial is a trinomial.