If ages are normally distributed then using a significance level of 0.05 , we can claim that the population mean is greater than 24.
Given sample mean of 25 years , sample size of 16 ,standard deviation of 2 years.
We have to find whether the population average is greater than 24.
We have to use t test because n is less than 30. It is right tailed.
We have to first form Hypothesis.
:μ>24,
:μ<24.
t=X-μ/S/
t critical at 5% significance level and degree of freedom=15 (16-1)=1.7531
t=24-25/2/
=-1/0.5
=-2
Because 1.7531 is greater than -2 so we will accept the null hypothesis.
Hence we can say that the average of population is greater than 24.
Learn more about t test at brainly.com/question/6589776
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Compatible numbers is used to find estimate to a sum. The numbers are chosen because its 'simplicity' when they are multiplied, divided, added or subtracted.
Estimate one:
We can decrease the value 1322 to 1320 and 18 to 15
Then do the sum = 1320÷15 = 88
This estimate is not far from the actual answer 1322÷18 = 73.44..
Estimate two:
We can increase 1322 to 1330 and 18 to 20
Then do the sum 1330÷20=66.50
This estimate is also not far from the actual answer 73.44...
Answer:
a = 8
Step-by-step explanation:
a ÷ 2 - 8 + 8 = - 4 + 8
a ÷ 2 = 4
a ÷ 2 × 2 = 4 × 2
a = 8
We have the function f(x)=(x-3)^2-49.
This equals...
0 = (x-3)^2 - 49
Add 49.
49 = (x-3)^2
Square both sides.
<span>±7 = x-3
</span><span>± 10 = x</span>
