Answer:
9 1/6
You may use a fraction calulator for this question by the way!
Hope this helps!
Sorry if its wrong!
Have a nice day!
Answer: a) (0.755, 0.925)
Step-by-step explanation:
Let p be the population proportion of 8th graders are involved with some type of after school activity.
As per given , we have
n= 100
sample proportion: 
Significance level : 
Critical z-value : 
(using z-value table)
Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-
i.e. 
i.e. 
i.e. 
Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)
To solve for the confidence interval for the population
mean mu, we can use the formula:
Confidence interval = x ± z * s / sqrt (n)
where x is the sample mean, s is the standard deviation,
and n is the sample size
At 95% confidence level, the value of z is equivalent to:
z = 1.96
Therefore substituting the given values into the
equation:
Confidence interval = 3 ± 1.96 * 5.8 / sqrt (51)
Confidence interval = 3 ± 1.59
Confidence interval = 1.41, 4.59
Therefore the population mean mu has an approximate range
or confidence interval from 1.41 kg to 4.59 kg.
The fraction of 200000.900 is <span>200,000 9/10 in mixed numbers. </span>
<span>Hope it helps!</span>
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Answer:
y = (1/2) cos (x)
Step-by-step explanation:
It has the same period as cos x but only goes half as high and half as low. Normalyy, cos(x) goes up to 1 and down to -1.
