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2 years ago

A halyard 12 ft long is cut into two pieces. Use the same variable to express the lengths of the two pieces.

Mathematics

1 answer:

Anarel [89]2 years ago

4 0

Answer:

The length is 6ft if cut in half because 6+6=12 which means 6ft+6ft=12ft

Step-by-step explanation:

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Determine the perimeter of a rectangle that is 10 inches long and 9 inches wide.<br> inches

Alex777 [14]

Answer:

Step-by-step explanation:

2(l + b)

2(10 + 9)

2(19)

2*19

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2 years ago

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What is the combined weight of all the kittens

hram777 [196]

Answer:

$5$

Step-by-step explanation:

3/4+12/8+3/2+10/8=5

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3 years ago

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Zahra gave out a survey to some students in her school about their favorite color. Students could choose between purple and red.

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2 years ago

You pick one card from a standard deck of 52 cards. What is the probability that the card will be a two?

Natali [406]

Answer:

4 out of 52 or simplified 2 out of 26

Step-by-step explanation:

There are 4 suites of cards, spades, hearts, clubs, diamonds and there is one card for two in each suite, so we therefore know that the probability of picking a two out of a 52 card deck will be four out of 52 or the simplified answer would be 2 out of 26. Hope this helps, have a great day!

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3 years ago

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

hram777 [196]

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.

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3 years ago