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Lady bird [3.3K]
3 years ago
5

The mass of an ant is 1/10 of the load, which it can carry in one try. What is the mass of the ant, if in one time it can carry

7/250 gram load?
Mathematics
1 answer:
Viefleur [7K]3 years ago
8 0

If you call m the mass of the ant and l the load, we have the equation

m = \cfrac{l}{10}

In fact, the mass of the ant is one tenth of the load, which is exactly what this equation states.

Since we are given the load, we simply need to plug its value in the equation to deduce the mass of the ant:

m = \cfrac{\frac{7}{250}}{10} = \cfrac{7}{250}\cdot\cfrac{1}{10} = \cfrac{7}{2500}

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Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one anothe
scZoUnD [109]

Answer:

a) P(male=blue or female=blue) = 0.71

b) P(female=blue | male=blue) = 0.68

c) P(female=blue | male=brown) = 0.35

d) P(female=blue | male=green) = 0.31

e) We can conclude that the eye colors of male respondents and their partners are not independent.

Step-by-step explanation:

We are given following information about eye colors of 204 Scandinavian men and their female partners.

              Blue    Brown     Green    Total

Blue        78         23            13          114

Brown     19         23            12          54

Green     11           9             16          36

Total      108       55            41          204

a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?

Using the addition rule of probability,

∵ P(A or B) = P(A) + P(B) - P(A and B)

For the given case,

P(male=blue or female=blue) = P(male=blue) + P(female=blue) - P(male=blue and female=blue)

P(male=blue or female=blue) = 114/204 + 108/204 − 78/204

P(male=blue or female=blue) = 0.71

b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=blue) = 78/114

P(female=blue | male=blue) = 0.68

c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=brown) = 19/54

P(female=blue | male=brown) = 0.35

d) What is the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=green) = 11/36

P(female=blue | male=green) = 0.31

e) Does it appear that the eye colors of male respondents and their partners are independent? Explain

If the following relation holds true then we can conclude that the eye colors of male respondents and their partners are independent.

∵ P(B | A) = P(B)

P(female=blue | male=brown) = P(female=blue)

or alternatively, you can also test

P(female=blue | male=green) = P(female=blue)

P(female=blue | male=blue) = P(female=blue)

But

P(female=blue | male=brown) ≠ P(female=blue)

19/54 ≠ 108/204

0.35 ≠ 0.53

Therefore, we can conclude that the eye colors of male respondents and their partners are not independent.

7 0
3 years ago
Find the lcd of the pair of expressions
STALIN [3.7K]

\frac{2}{7}  {x}^{10}  {y}^{4}  \: and \:  \frac{2}{14}  {x}^{8}  {y}^{7} \\ lcd = 14 {x}^{10}   {y}^{7}
Answer is
14 {x}^{10}  {y}^{7}

When it's powered numbers, remember to choose the one with the largest power. For example,
{x}^{10}
And
{x}^{8}
Choose 10, same goes to y, where you choose 7 not 4 since 7 is larger.

Hope this helps. - M
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What is this fraction as a decimal?
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Answer:

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