7/15 will be the most you can simplify
Given that,
Sample size= 83
Mean number= 39.04
Standard deviation= 11.51
We know the critical t-value for 95% confidence interval which is equal to 1.989.
We also know the formula for confidence interval,
CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))
So, we have
CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)
CI= (39.04 - 2.513,39.04 + 2.513)
CI= (36.527,41.553)
Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).
Answer:
260 square yards
Step-by-step explanation:
(9+17/2)*20
Fter<span> a </span>125<span>% </span>markup<span> and a </span>10<span>% </span>discount<span> the </span>price<span> of a </span>watch<span> is 30.78 </span>before tax<span> what was the </span>wholesale price<span> - 2929721. ... If an item is $15.20 is increased by </span>125<span>% (34.20) it would be about $34.20 </span>before<span> we subtract </span>10<span>% (3.42)....we arrive at the end </span>cost<span> of </span>$30.78<span>. </span>wholesale price<span>would be 15.20.</span>
Answer:
-2x³+4x
Step-by-step explanation:
g(x) + f(x)
(2x) + (-2x³+2x)
-2x³+4x
