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kvasek [131]
3 years ago
12

Please answer this !

Mathematics
1 answer:
AURORKA [14]3 years ago
8 0

Answer:

50

Step-by-step explanation:

Using the experimental probability, first we have to find total amount paint then to find the amount of good quality forever paint we have to subtract the amounts of other paints from total amount.

Let x represents the amount of good quality forever paint sold.

Total amount of paint =48+45+35+39+33+x=200+x     ...(i)

Experimental probability of paint sold will be better quality durable paint is

P=\dfrac{\text{Amount of better quality durable paint}}{\text{Amount of total paint}}

P=\dfrac{45}{\text{Amount of total paint}}

It is given that experimental probability of paint sold will be better quality durable paint is \frac{9}{50}.

\frac{9}{50}=\dfrac{45}{\text{Amount of total paint}}

\text{Amount of total paint}=\dfrac{45\times 50}{9}

\text{Amount of total paint}=250     ...(ii)

From (i) and (ii), we get

200+x=250

x=250-200

x=50

Therefore, 50 gallons good quality forever paint is sold.

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Thank you!!!!!!!!!!!!!!!
OverLord2011 [107]

Answer:

Binomial distribution requires all of the following to be satisfied:

1. size of experiment (N=27) is known.

2. each trial of experiment is Bernoulli trial (i.e. either fail or pass)

3. probability (p=0.14) remains constant through trials.

4. trials are independent, and random.

Binomial distribution can be used as a close approximation, with the  usual assumption that a sample of 27 in thousands of stock is representative of the population., and is given by the probability of x successes (defective).

P(x)=C(N,x)*p^x*(1-p)^(n-x)

where N=27, p=0.14, and C(N,x) is the number of combinations of x items out of N.

So we need the probability of <em>at most one defective</em>, which is

P(0)+P(1)

= C(27,0)*0.14^0*(0.86)^(27) + C(27,1)*0.14^1*(0.86^26)

=1*1*0.0170 + 27*0.14*0.0198

=0.0170+0.0749

=0.0919


4 0
3 years ago
Solve this equation! Please Help!<br><br> √3 x 66.15/4.41<br><br> (SHOW YOUR WORK)
finlep [7]
Hi Jessica,
<span>Remember PEMDAS (Parenthesis, Exponents, Multiplication & Division, Addition & Subtraction).

√3 x 66.15/4.41 {Exponents/Cube & Square Roots First}

1.73 x 66.15 ÷ 4.41 {Multiplication}

114.4395 ÷ 4.41 {Division}

25.95 {Final Answer}

Cheers,
Izzy</span>
7 0
3 years ago
Find the value of x.
VladimirAG [237]

mmdmdkk d vsysosozbhs..x

s.jskduid

zjsjshueks

shsbks8s s

gsoshtsmsmysm

<h3>ts</h3><h3>syw</h3>

Step-by-step explanation:

<h2><em><u>jabhshjannsbsb</u></em><em><u> </u></em><em><u>zbzgiwbztsksbztqns</u></em><em><u>.</u></em><em><u>hsjwj9sjysyqjzguwjshuwiqjys71jsy5169191ywbs</u></em></h2><h3><em><u>qhqk9qj.qhqikqhj2nkqjksuwkqjsuwkjswjj</u></em><em><u> </u></em><em><u>w</u></em><em><u> </u></em><em><u>y</u></em><em><u> </u></em><em><u>www</u></em><em><u> </u></em><em><u>kwb8wkjuwbshsh8wjsoskhjqhs8kshzkwns</u></em></h3>
3 0
3 years ago
Can someone help me ASAP??? its for homeworkkk
r-ruslan [8.4K]

I think that the answer is 120 cubes

6 0
3 years ago
(5 points) An urn contains two blue balls denoted by B1 and B2, and three white balls denoted by W1, W2 and W3. One ball is draw
never [62]

Answer:

The probability of the event that first ball that is drawn is blue  is \frac 25.

Step-by-step explanation:

Probability:

If S is is an sample space in which all outcomes are equally likely and E is an event in S, then the probability of E,denoted P(E) is

P(E)=\frac{\textrm{The number of outcomes E}}{\textrm{The total number outcomes of S}}

Given that,

An urn contains two balls B₁ and  B₂ which are blue in color and W₁,W₂ and W₃ which are white in color.

Total number of ball =(2+3) =5

The number ways of selection 2 ball out of 5 ball is

=5²

=25

Total outcomes = 25

List of all outcomes in the event that the first ball that is drawn is blue are

B₁B₁ , B₁B₂ , B₁W₁ , B₁W₂ , B₁W₃ , B₂B₁ , B₂B₂ ,  B₂W₁ , B₂W₂ , B₂W₃

The number of event that the first ball that is drawn is blue is

=10.

The probability of the event that first ball that is drawn is blue  is

=\frac{10}{25}

=\frac25

3 0
3 years ago
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