To solve this
problem, let us analyze this step by step. The temperature for each day is as
follows:
Water temperature
on Sunday = 78 degrees F
Water temperature
on Monday = changed by -3 degrees F
Water temperature
on Tuesday = changed by 3 degrees F
We can see that
the total change of water temperature from Sunday to Tuesday is:
-3 + 3 = 0
Therefore there
is zero overall change. There the integer which represents the temperature
change is “0”.
Since the overall
change in water temperature is zero, hence the temperature on Sunday and on
Tuesday is similar.
Water temperature
on Tuesday = 78 degrees F
Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: the amount a 10-11-year-old spends on a trip to the mall.
Assuming that the variable has a normal distribution, you have to construct a 98% CI for the average of the amount spent by the 10-11 year-olds on one trip to the mall.
For this you have to use a Student-t for one sample:
X[bar] ± *
n= 6
$18.31, $25.09, $26.96, $26.54, $21.84, $21.46
∑X= 140.20
∑X²= 3333.49
X[bar]= ∑X/n= 140.20/6= 23.37
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/5[3333.49-(140.2)²/6]= 11.50
S= 3.39
X[bar] ± *
[23.37 ± 3.365 * ]
[18.66;29.98]
With a confidence level of 98%, you'd expect that the interval $[18.66;29.98] will include the population mean of the money spent by 10-11 year-olds in one trip to the mall.
I hope you have a SUPER day!
Answer:
it's either B or D but you would have to do the math
Step-by-step explanation:
F(x)<span>= -10
</span>f(5)=-10
f=-10 divided by 5
f=-2