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hjlf
3 years ago
6

A study addressed the issue of whether pregnant women can correctly predict the gender of their baby. Among 104 pregnant​ women,

57 correctly predicted the gender of their baby. If pregnant women have no such​ ability, there is a 0.327 probability of getting such sample results by chance. What do you​ conclude?
Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

Hence the study seem to be inconclusive since the probability of getting those result by chance are somewhat very high.

Step-by-step explanation:

The result seem to be inconclusive since the probability of getting those result by chance are somewhat very high.

Since the probability of getting such sample result by chance are very high, we cannot assume that the women have the ability to predict the gender of their babies.

Hence the study seem to be inconclusive since the probability of getting those result by chance are somewhat very high

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romanna [79]
Together it would take 3 hours.
Team A can wrap all gifts in 7 hours; thus they can wrap 1/7 of the gifts in 1 hour.

Team B can wrap all gifts in 5 hours; thus they can wrap 1/5 of the gifts in 1 hour.

1/7h + 1/5h = 1, where h is the number of hours; it equals 1 because it is 100% of the gifts;

Find a common denominator.  35 is the first thing 7 and 5 will both divide into.  Convert the fractions:
5/35h + 7/35h = 1
12/35h = 1

Divide both sides by 12/35:
12/35h ÷ 12/35 = 1 ÷ 12/35
h = 1/1 ÷ 12/35
h = 1/1 × 35/12
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4 0
3 years ago
The width of a rectangular garden is 4.5 meters shorter than its length which is x meters.
Contact [7]

Answer:

  A.  width = x - 4.5

  B.  111 = 2(x + (x -4.5))

Step-by-step explanation:

A. The width is 4.5 meters shorter than x, so is x -4.5.

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B. The perimeter is twice the sum of length and width. It is 111 meters, so an equation for it could be ...

  111 = 2(x + (x -4.5))

4 0
4 years ago
How can i prove this property to be true for all values of n, using mathematical induction.
chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

\implies \: 1  =  {1}^{2}  =  \frac{1(1 + 1)(2 + 1)}{6}

\implies{ \frac{1(2)(3)}{6} }

\implies{ 1}

Therefore, it is true for the first part.

In the second part we will assume that,

\: {  {1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  =  \frac{k(k + 1)(2k + 1)}{6}  }

and we will prove that,

\sf{ \: { {1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} =  \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}

\: {{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2}  =  \frac{(k + 1)(k + 2) (2k + 3)}{6}}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} +  \frac{(k + 1) ^{2} }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}

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{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}

<u>Henceforth, by </u><u>using </u><u>the </u><u>principle </u><u>of </u><u> mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n</u>.

_______________________________

<em>Please scroll left - right to view the full solution.</em>

8 0
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neonofarm [45]

Answer:

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5 0
3 years ago
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