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Ymorist [56]
3 years ago
14

HIII ANOTHER ONE last one was blurry. ty guys!!

Mathematics
1 answer:
nikklg [1K]3 years ago
5 0

Answer:

The second one

Step-by-step explanation:

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A random sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the
tester [92]

Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:

<span>CI = X ± t * s / sqrt(n)                      ---> 1</span>

Where,

X = the sample mean = 84

t = the t score which is obtained in the standard distribution tables at 95% confidence level

s = sample variance = 12.25

n = number of samples = 49

From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:

t = 1.68

 

Therefore substituting the given values to equation 1:

CI = 84 ± 1.68 * 12.25 / sqrt(49)

CI = 84 ± 2.94

CI = 81.06, 86.94

 

<span>Therefore at 95% confidence level, the scores is from 81 to 87.</span>

6 0
3 years ago
Find the missing measure.
eduard

Answer:

59

Step-by-step explanation:

(360-92-150)/2

= 118/2

= 59

Answered by GAUTHMATH

4 0
3 years ago
A 25-pound bag of bird food costs $19.50.A 30-pound bag of the same bird food costs $5.18.A 26-ounce opackage costs $10.40. Whic
Lyrx [107]
The 30 pound bag is the bust guys it's bigger and same price and in the long run would be better
7 0
3 years ago
Read 2 more answers
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
Elapsed time in minutes<br> Start time 3:10pm <br> end time 4;00pm
gizmo_the_mogwai [7]
50 minutes has elapsed.  
6 0
3 years ago
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