Can someone help me with this question it’s 7.55 x 4.5
2 answers:
If you are going to multiply this by hand, make sure that you add a zero to the end of 4.5 so that it is 4.50. Because there are four decimal places, you need to make sure your answer has four decimal places as well. Then add a row of zeroes to the bottom because anything times zero equals zero. Next, add a zero in the hundredths place and multiply 5 from 4.50 to 7.55. Then, do the same with the 4. Add all your numbers together and you should get 33.9750 or 33.975.
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Answer:
a) 0.259
b) 0.297
c) 0.497
Step-by-step explanation:
To solve this problem, it is important to know the normal probability distribution and the central limit theorem.
Normal probability distribution
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.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.6293.
X = 10.8
has a pvalue of 0.3707.
0.6293 - 0.3707 = 0.2586
0.259 probability, rounded to three decimal places.
b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?
Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So
X = 11
has a pvalue of 0.5.
X = 10.5
has a pvalue of 0.2033.
0.5 - 0.2033 = 0.2967
0.297, rounded to three decimal places.
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.7486.
X = 10.8
has a pvalue of 0.2514.
0.7486 - 0.2514 = 0.4972
0.497, rounded to three decimal places.
M∠A = 42° [isosceles triangle]
m∠B = 180 - 42 - 42 = 96° [in a triangle, the three interior angles always add to 180°]
m∠C = 180 - 96 - (x+12)
m∠C = 84 - x - 12
m∠C = 72 - x
2x+9 + 3x-1 + 72-x = 180
4x + 80 = 180
4x = 180 - 80
4x = 100
x = 100/4
x = 25
m∠C = 72 - x = 72 - 25 = 47°
<span>I hope this helped</span>
m∠B = 180 - 42 - 42 = 96° [in a triangle, the three interior angles always add to 180°]
m∠C = 180 - 96 - (x+12)
m∠C = 84 - x - 12
m∠C = 72 - x
2x+9 + 3x-1 + 72-x = 180
4x + 80 = 180
4x = 180 - 80
4x = 100
x = 100/4
x = 25
m∠C = 72 - x = 72 - 25 = 47°
<span>I hope this helped</span>