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Sladkaya [172]
3 years ago
13

Imagine that you are conducting a one-variable chi-square test to investigate the hypothesis that there are equal numbers of cat

lovers and dog lovers in the office at work. Having conducted a survey, you found 150 preferred dogs and 120 preferred cats. What would the expected frequencies be in each cell? 135 150 and 120 270 More information is needed to calculate the expected frequencies.
Mathematics
1 answer:
Talja [164]3 years ago
4 0

Answer:

135

Step-by-step explanation:

Based on the following information:

- There are 150 dog lovers

- There are 120 cat lovers

So:

The null hypothesis:

H, there are an equal number of dog lovers and cat lovers, so the expected frequency in each cell will be the same and that is:

f = (150 + 120) / 2

f = 270/2 = 135

then the first option of 135 is correct

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There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0
3 years ago
Is 11/12 greater or less than 2/3?
Sauron [17]
False beacause 2/3 is bigger than 11/12. Think about it like pizza 2/3 pieces is better cuz there bigger & 11/12 are smaller pieces.
8 0
3 years ago
The measure of the supplement of an angle is seven times as large as the measure of its complement find the measure of the angle
8090 [49]

Answer: original angle = 75°

Its supplement 105°

Its complement = 15°

Step-by-step explanation:

We know that,

  • Sum of supplementary angles is 180 degrees.
  • Sum of complementary angles is 90 degrees.

Let x be the original angle , then its supplement = 180° - x

its complement = 90° - x

As per given,

180° - x = 7(90° - x)

⇒ 180° - x = 630° - 7x

⇒  7x- x = 630° -180°

⇒  6 x = 450°

⇒  x = 75°  [divide both sides by 6]

So, original angle = 75°

Its supplement =  180° - 75° = 105°

Its complement = 90° -75° = 15°

Hence,  original angle = 75°

Its supplement 105°

Its complement = 15°

4 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
2 years ago
What is the area of 16 inches and 3 feet in square meters
Artyom0805 [142]
You just multiply 16 and 3 and it gives you 58
3 0
3 years ago
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