Step-by-step explanation:
expanded = 3n + 18
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Answer:
a) 99.97%
b) 65%
Step-by-step explanation:
• 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
• 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
• 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
Mean of 98.35°F and a standard deviation of 0.64°F.
a. What is the approximate percentage of healthy adults with body temperatures within 3 standard deviations of the mean, or between 96.43°F and 100.27°F?
μ - 3σ
98.35 - 3(0.64)
= 96.43°F
μ + 3σ.
98.35 + 3(0.64)
= 100.27°F
The approximate percentage of healthy adults with body temperatures is 99.97%
b. What is the approximate percentage of healthy adults with body temperatures between 97 .71°F and 98.99°F?
within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
μ - σ
98.35 - (0.64)
= 97.71°F
μ + σ.
98.35 + (0.64)
= 98.99°F
Therefore, the approximate percentage of healthy adults with body temperatures between 97.71°F and 98.99°F is 65%
Her answer represents "-3.7 less than 4 times a number is 11.9". The correct equation should be -3.7n-4=-11.9.
Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
The figure below shows a diagram of this problem. First of all we graph the hemisphere. This one has a radius equal to 1. Given that z ≤ 0 a sphere will be valid only in the negative z-axis, that is, we will get a half of a sphere that is the hemisphere shown in the figure. We know that this hemisphere is oriented by the inward normal pointing to the origin, then we have a Differential Surface Vector called
N, using the Right-hand rule <span>the boundary orientation is </span>counterclockwise.
Therefore, the answer above
False