<span>Minimal residual cancer cells may not be detected in surgical margins of oral squamous cell carcinoma (OSCC) with routine histological examination. Using molecular markers at surgical margins can be helpful. We attempted to evaluate the MMP-9 and E-cadherin expression in OSCC samples and tumor-free surgical margins and association with clinicopathological factors. We examined E-cadherin and MMP-9 expression in 58 OSCCs including 19 grade I, 21 grade II and 18 grade III with histological tumor-free surgical margins by immunohistochemistry. Specimens were also divided in two groups: 19 samples as an early and 39 as an advanced stage. For E-cadherin in OSCCs and surgical margins, significant difference was observed between poor and moderate tumor differentiation. Different stages of OSCC demonstrated significant differences with higher expression in early stage tumors. For surgical margins, 82.1% of advanced and 84.2% of early stage samples demonstrated immunoreactivity. Both OSCC samples and surgical margins demonstrated significant differences for MMP-9 between stages with higher immunoreactivity in advanced stage, whereas there were not differences between different grades in surgical margins. E-cadherin and MMP-9 expression at histologically negative surgical margins shows the significance of these markers for prognostic values in OSCC patients with E-cadherin being the preferred predictor.</span>
Answer:
858 thousand pounds
Step-by-step explanation:
The first field has an area of 16 acres; the second has an area "1 1/4 times bigger".
Most people will interpret that incorrectly, saying the area of the second field is 16 time 1 1/4, which is 20. But the grammatically correct interpretation of "1 1/4 times bigger" is the size of the first field, plus 1 1/4 more times the size of the first field.
So the size of the second field in acres is 16+%2B+1.25%2A16+=+36
So the first field has an area of 16 acres and the second has an area of 36 acres.
The first field yields 37.5 thousand pounds of wheat per acre. The total amount of wheat from the first field is 16%2A37500+=+600000
The yield from the second filed is 1 2/15 times as much per acre as the first field. The yield in pounds per acre for the second field is 37500%2A%2817%2F15%29+=+40500.
So the total amount of wheat from the second field is 36%2A40500+=+1458000
The difference in the total yield from the two fields is 1458000-600000+=+858000
The second field yielded 858 thousand pounds more of wheat.
Answer:
The probability that x is less than 9.7 is 0.0069 = 0.69%
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:
The probability that x is less than 9.7 is _____.
This is the pvalue of Z when X = 9.7. So
has a pvalue of 0.0069
The probability that x is less than 9.7 is 0.0069 = 0.69%
