Answer:
14.63% probability that a student scores between 82 and 90
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that a student scores between 82 and 90?
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So
X = 90
has a pvalue of 0.9649
X = 82
has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90
Answer:
16 cm2
Step-by-step explanation:
By equa tion of area of triangle: S = 1/2 * height * base
Triangle B: 32 = 1/2 * 8 * x --> x = 8 (cm)
The base of triangle A is half of the base of triangle B so it is 4 cm.
The area of triangle A = 1/2 * 8 * 4 = 16 (cm2)
Answer:
25cm
Step-by-step explanation:
one the possibility 36 - 11 =25 the answer is 25
Answer:
Test statistic = 1.3471
P-value = 0.1993
Accept the null hypothesis.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4
Sample mean, 
= 4.8
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation, s = 2.3
First, we design the null and the alternate hypothesis
We use two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Now, we calculate the p-value.
P-value = 0.1993
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.
4. 15 tens, 1 hundred, 5 tens
150
5. 18 tens, 1 hundred, 8 tens
180
note: I hope this helps. I haven't done this in a long time. Sorry X(
