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7nadin3 [17]
3 years ago
9

50% of students change their major area of study after their first year in a program. A random sample of 100 students in the Col

lege of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program
Mathematics
1 answer:
devlian [24]3 years ago
4 0

Answer:

Hence, since z ( - 0.4 ) is greater than ( -1.65 );

we reject H₀, Null hypothesis,

Therefore, There is no evidence that there has been a significance decrease in the proportion of students who change their major after the first year in the program.

Step-by-step explanation:

Given the data in the question;

Let p = 50% = 0.5

Hypothesis;

Null hypothesis            H₀ : p ≥ 0.5

Alternative hypothesis H₁ : p < 0.5

x = 48

sample size n = 100

sample proportion p" = x / n = 48 / 100 = 0.48

significance value = 0.5

Z_\alpha = -1.65

Test Statistics;

z = (p" - p) / √( p( 1 - p ) / n )

we substitute

z = (0.48 - 0.5) / √( 0.5( 1 - 0.5 ) / 100)

z = -0.02 / √( 0.25 / 100 )

z = -0.02 / √0.0025

z = -0.02 / 0.05

z = -0.4

Hence, since z ( - 0.4 ) is greater than ( -1.65 );

we reject H₀, Null hypothesis,

Therefore, There is no evidence that there has been a significance decrease in the proportion of students who change their major after the first year in the program

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Answer:

see explanation

Step-by-step explanation:

Using the exact values of the trigonometric ratios

sin60° = \frac{\sqrt{3} }{2}, cos60° = \frac{1}{2}

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Using the sine ratio on the right triangle on the left

sin60° = \frac{opposite}{hypotenuse} = \frac{a}{4\sqrt{3} } = \frac{\sqrt{3} }{2}

Cross- multiply

2a = 4\sqrt{3} × \sqrt{3} = 12 ( divide both sides by 2 )

a = 6

Using the cosine ratio on the same right triangle

cos60° = \frac{adjacent}{hypotenuse} = \frac{c}{4\sqrt{3} } = \frac{1}{2}

Cross- multiply

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c = 2\sqrt{3}

------------------------------------------------------------------------------------------

Using the sine/cosine ratios on the right triangle on the right

sin45° = \frac{a}{b} = \frac{6}{b} = \frac{1}{\sqrt{2} }

Cross- multiply

b = 6\sqrt{2}

cos45° = \frac{d}{b} = \frac{d}{6\sqrt{2} } = \frac{1}{\sqrt{2} }

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a = 6, b = 6\sqrt{2}, c = 2\sqrt{3}, d = 6

3 0
3 years ago
Help: I am a number between 60 and 100. My ones digit is six less than my tens digit. I am a prime number.
Reptile [31]

Answer:

Let's name the digit:

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y - tens digit

we know that x=y-2.

Now, y can be 6,7,8,9 (the number is between 60 and 100 (so depending on your understanding of "between", 0 is also possible. but then the number would have to have -2 as its ones digit, so in any case, it's not possible).

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Out of those 64 and 86 are even, so they can't be prime.

75 has 5 in its ones number: it's divisible by 5.

so the correct answer is 97.

Step-by-step explanation:

Hope this helps!

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