Answer:
M=2,179.29
The average exceeds the estimate
Her expenses would need to be $1,625.00
Step-by-step explanation:
1.) What is the mean of her friends' personal expenses?
The mean is defined as the average expense value, calculated as follows:
2.) Is the average higher or lower than the estimate?
Given the estimate of $2,110, the average is indeed higher than this estimate.
3.)What amount of personal expenses would Ashley need to maintain a combined average of $2,110 with her friends' amounts for that school year?
Her needed amount of expenses would be $1,625.00.
We determine that the true average calorie content as estimated in the sampled population surpasses the actual calorie content. Step-by-step explanation: An article discussed a pilot study where each of the 58 participants was asked to estimate the calorie count of a 12 oz beer known to have 153 calories. The observed sample mean of calorie estimation was 193, with a sample standard deviation of 88. Let
=
true average estimated calorie level within the sampled population. Thus, Null Hypothesis, : 153 calories {indicating that the true average estimated calorie content does not exceed the actual amount}. Alternative Hypothesis, 
: > 153 calories {indicating the true average estimation exceeds the actual}. The appropriate test statistic would be a one-sample t-test statistic, as we lack knowledge of the population standard deviation; Test Statistic = ~t = 
where, sample mean estimated calorie level = 193 calories, s = sample standard deviation = 88, and n = sample size = 58. Therefore, the test statistic = ~t = 3.462. The t-table indicates a critical value of 1.6725 for 57 degrees of freedom at a 0.05 significance level. Since our test statistic of 3.462 > 1.6725, we have sufficient evidence to reject the null hypothesis; thus, affirming that the true average estimated calorie content in the sampled population exceeds the real content.
The given expression is
Thus, in verbal form, we can express it as one eighth of a number.
Answer:
Upper limit: 3.116
Lower limit: 3.125
Step-by-step breakdown:
The random circle drawn by Hellen has a circumference of C=405 mm, measured to 3 significant figures, leading to a diameter, d, of 130 mm, accurate to 2 significant figures.
We utilize the formula
Substituting in the figures provides:
Consequently, Hellen’s value for π is recorded as 3.115 to three decimal places.
To find both limits, we divide the specified level of precision by two.
This figure is then added to the rounded number to determine the upper limit:
3.115+0.0005=3.1155
The lower limit calculates as 3.115-0.0005=3.1145
