Since this is a ratios and proportions problem, you would set up the equation like this:
1/50,000 = x/8 x is the length of the distance on the map
After that, you cross multiply and you will get 50,000x=8
You then solve for x to get 0.00016
we know that
The probability that "at least one" is the probability of exactly one, exactly 2, exactly 3, 4 and 5 contain salmonella.
The easiest way to solve this is to recognise that "at least one" is ALL 100% of the possibilities EXCEPT that none have salmonella.
If the probability that any one egg has 1/6 chance of salmonella
then
the probability that any one egg will not have salmonella = 5/6.
Therefore
for all 5 to not have salmonella
= (5/6)^5 = 3125 / 7776
= 0.401877 = 0.40 to 2 decimal places
REMEMBER this is the probability that NONE have salmonella
Therefore
the probability that at least one does = 1 - 0.40
= 0.60
the answer is
0.60 or 60%
Answer:
26.11% of the test scores during the past year exceeded 83.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by
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:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that 
.
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when 
. So:
has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.
Answer:
Area of drawing= 0.315 m²
Step-by-step explanation:
Base = 6m
Height = 4.2 m
Area = height* base
Area= 4.2*6
Area= 25.2 m²
The scale is 10 cm on the drawing for every 8 meters on the sculpture.
Which is 10 cm to 800 cm
Or
0.1m to 8m
Arra of sculpture= 25.2 m²
area of drawing= (25.2*0.1)/8
Area of drawing= 0.315 m²
