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irga5000 [103]
3 years ago
10

To the nearest square unit, what is the area of the regular heptagon shown below?

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
4 0

Answer:

B. 1783 square units

Step-by-step explanation:

user100 [1]3 years ago
3 0

Answer:

Its b: 1572 square units

Step-by-step explanation:

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What is the unit form for 10 x 3tens
Tems11 [23]

Answer:

3 hundreds

Step-by-step explanation:

10*3*10=300

unit for is 3 hundreds standard form is 300

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3 years ago
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Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t
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Answer:

a.\mu=15

b.\mu=7.8586 \ and  \ \mu=22.1414

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

P(X=k)=C_k^n.p^k.(1-p)^{n-k}

Let n be the number of trials,n=100

and p be the probability of success,p=15\%

The mean of a binomial distribution is the probability x sample size.

\mu=np=100\times0.15=15

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as:\sigma=\sqrt npq\\q=1-p

Therefore, \sigma=\sqrt(100\times0.015\times0.85)=3.5707

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c.  x=45 is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0
3 years ago
The annual 2-mile fun-run is a traditional fund-raising event to support local arts and sciences activities. It is known that th
victus00 [196]

Answer: 0.0228

Step-by-step explanation:

Given : The  mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

\mu=30\\\\\sigma=5.5

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Let X be the random variable that represents the finish times for this event.

z score : z=\dfrac{x-\mu}{\sigma}

z=\dfrac{19-30}{5.5}=-2

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

P(X

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228

8 0
3 years ago
When is a lower annual interest fee better than a low annual fee
sertanlavr [38]

hope it helps.........

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3 years ago
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A hyena can run one over four of a mile in 22.5 seconds. At this rate, which equation can be used to determine how fast a hyena
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he runs 98 mph

Step-by-step explanation:

Sorry if im wrong

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