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PolarNik [594]
3 years ago
9

Leciały kaczki: jedna na przedzie, dwie za nią, jedna z tyłu, dwie przed nią,jedna pomiędzy dwiema i trzy pod rząd. Ile kaczek l

eciało?
Potrzebne na teraz !
Mathematics
1 answer:
balandron [24]3 years ago
7 0
I  think its 8, if you count up all the ducks it's 7. But it's talking like it's counting the ducks in front,middle and behind a duck so 7+1=8.
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Miss Trejo ran 120 yard. What are equivalent distance to how far she ran?
erastova [34]

Answer: 360 feet, or 3/44 of a mile

Step-by-step explanation: One yard is 3 feet, and 1760 yards are in a mile

7 0
2 years ago
I need the answer for number 3.
deff fn [24]

Answer:

0.12

Step-by-step explanation:

15 divided by 125 is equal to 0.12

6 0
3 years ago
High concentrations of carbon monoxide CO can cause coma and possible death. The time required for a person to reach a COHb leve
ale4655 [162]

Answer:

A) 10.38hours

B) 617.95 is the concerntration of CO

Step-by-step explanation:

A) when x=600

T= 0.0002x^2 - 0.316x + 127.9

T= 0.0002(600)^2 -0.316(600) + 127.9

T= 10.38secs

B) to find the concerntration, use Almighty formular

X=-b+-sqrt(b^2-4ac)/2a

0.0002x^2-0.316x-127.9=9

0.0002x^2 - 0.316x + 118.9=0

X= 0.316 +- sqrt(0.316^2) -4(0.0002×118.9)/(2×0.0002)

X= 0.316+(0.68817/0.0004)=962.05 or

X=-0.316 -0.68817/0.0004 =617.95

Therefore X is within the domain of 617.95 is the concerntration

5 0
3 years ago
Attend to precision. Suppose that 21 and 22 are same-side interior angles formed by two parallel lines cut by a transversal, and
Zepler [3.9K]

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5 0
2 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
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