1/6 +3/6 = 4/6 but if you simply its 2/3
Answer:
Step-by-step explanation:
So first you would find a common denominator. The common denominator is 6. For 1/3 you multiply both numbers by 2. For 1/2 you multiply both numbers by 3. You would then have 3/6 and 2/6. You add them together which equals 5/6. Timothy need 5/6 cups of bread crumbs for his casserole.
The final answer is the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than 143lb is 1.0000.
A normal distribution is a continuous data distribution with a bell-shaped curve. The normally distributed random variable X has mean
and standard deviation
.
In addition, the standard normal distribution represents a normal curve with a mean of zero and a standard deviation of one.
The number of standard deviations an element deviates from the mean is indicated by a standardized Z-score.
What is the theorem of the central limit?
- If the sample size from the population is sufficiently large and has a finite variance, the mean of all samples taken will be approximately the same as the population mean. And the variance would be the population variance divided by the sample size. In addition, the sampling distribution would be roughly normal.
- The formula for sampling distribution of the sample mean is

- The formula for standard deviation of the sample mean is,

- Here,
is the population standard deviation and n is the sample size. - The formula for standard z-score is,

- From the given information,Population mean,
,Population standard deviation,
,Sample size, n=70 - Compute the probability that the boat is overloaded because the mean weight of the 70 passengers is greater than
.

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Midpoint of (x1,y1) and (x2,y2) is
((x1+x2)/2,(y1+y2)/2)
just average the x and y values
(5,-8) and (10,18)
((5+10)/2,(-8+18)/2)
(15/2,10/2)
(7.5,5)
the midpoint is (7.5,5)
Answer:


Step-by-step explanation:
For this case we have the following info given:
Treatment: 12 13 15 19 20 21 24
Control: 18 23 24 30 32 35 39
We can find the sample mean and deviations with the the following formulas:


And repaplacing we got:
the sample mean for treatment
the sample mean for treatment
the sample deviation for treatment
the sample deviation for control
the sample size for each sample
The degrees of freedom are given by:

The confidence interval for the difference of means is given by:
The confidence is 98% so then the significance is
and
. Then the critical value would be:

And replacing we got:

