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bixtya [17]
3 years ago
7

A missile protection system consists of n radar sets operating independently, each with probability 0.9 of detecting a missile e

ntering a zone that is covered by all of the units.a) Find a nice expression for the probability of detecting at least one of the sets detects the missile. (Tip: what is the probability, that a missile is not detected by any of the sets? - and what has this probabilityto do with the above?)b) How large must n be if we require that the probability of detecting a missile that enters the zone be 0.999 ? 0.999999 ?c) If n = 5 and a missile enters the zone, what is the probability that exactly 4 sets detect the missile?At least one set?
Mathematics
1 answer:
stepladder [879]3 years ago
6 0

Answer:

A.) The probability that the missile is not detected by any of n sets = (1-0.9)^n = 0.1^n.

Where it's opposite will be that the missile is detected by atleast 1 of the sets. Which will be equals to 1-(0.1)^n.

B.) 0.999= 1-(0.1)^n

Here n=3

For, 0.999999=1-(0.1)^n

n=6.

C.) If n=5

5C4*(0.9)^4*(0.1)^1= 0.32805

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F(x) = x5 + (x + 3)2 x=-1
sveticcg [70]
Hello!

You put -1 in for x

F(-1) = (-1)^5 + (-1 + 3)^2

F(-1) = -1 + (2)^2

F(-1) = -1 + 4

F(-1) = 3

The answer is 3

Hope this helps!
8 0
3 years ago
Use the relationship between the angles in the figure to answer
Sholpan [36]
31+40 = 71. 180-71=109
5 0
3 years ago
Read 2 more answers
You randomly draw a marble out of a bag that contains 20 total marbles. 12 of the marbles in the bag are blue.
Advocard [28]

Answer:

P(Blue)=0.60

Step-by-step explanation:

Probability: Probability\ of\ an\ event=\frac{favourable\ outcome}{total\ outcome}

Total\ marbles=20\\\\Hence\ total\ outcomes=20\\\\Blue\ marbles=12\\\\Hence\ favourable\ outcomes=12\\\\P(Blue)=\frac{12}{20}\\\\P(Blue)=\frac{2\times 6}{2\times 10}\\\\P(Blue)=\frac{6}{10}\\\\P(Blue)=0.60

6 0
3 years ago
Chi square Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, perc
Marat540 [252]

Answer:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

Step-by-step explanation:

The missing data for the random sample of 80 days between December and April with snowfall is:

 Month            Days

December         16

January             11

February            16

 March              18

  April                19

The Chi-square goodness of fit test would be used to determine whether the snow falls in Daphne's hometown followed the given distribution.

The hypothesis for the test can be defined as follows:

<em>H</em>₀: The snow falls in Daphne's hometown does not follow the given distribution.

<em>Hₐ</em>: The snow falls in Daphne's hometown followed the given distribution.

Assume that the significance level of the test is, <em>α</em> = 0.05.

The Chi-square test statistic is given by:

\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}

The value of Chi-square test statistic is computed in the Excel sheet below.

\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}=8.383

Compute the <em>p</em>-value as follows:

\text{p-value}=P(\chi^{2}_{(4)}>8.383)=CHISQ.DIST.RT(8.383,4)=0.0785

The <em>p</em>-value of the test is 0.0785.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

<em>p</em>-value = 0.0785 > <em>α</em> = 0.05.

The null hypothesis will not be rejected.

Conclusion:

There is not enough evidence to suggest that the snow falls in Daphne's hometown does not follow the given distribution.

4 0
3 years ago
CAN YOU ANSWER THIS PLEASE
BabaBlast [244]
It’s the absolute value of 0 and 5 because it’s the same absolute value the absolute vale is 5 for both 5 and -5 because absolute value of a number is the distance from 0 you can’t have a negative value
5 0
3 years ago
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