Answer:
13 children and 9 adults if the total cost is $152.5
Step-by-step explanation:
Let x children and y adults
x + y = 22 (1)
5.5x + 9y = 125.5 (2)
y = 22 - x
5.5x + 9(22 - x) = 125.5
5.5x + 198 - 9x = 125.5
-3.5x = 125.5 - 198
-3.5x = -72.5
x = 20.7
y = 22 - x = 1.3
Which is not possible
If the total cost is $152.5
x + y = 22 (1)
5.5x + 9y = 152.5 (2)
y = 22 - x
5.5x + 9(22 - x) = 152.5
5.5x + 198 - 9x = 152.5
-3.5x = 152.5 - 198
-3.5x = -45.5
x = 13
y = 22 - 13 = 9
It could only be scalene or isosceles ... an equilateral triangle has all 60 degree angles
Isosceles- 90-45-45 degrees
Scalene- 90-35-55 degrees
The formula for getting the accumulated amount(compounded) is;
A =P(1+r%)∧n
Where A = Acumulated amount
P = principle (deposit)
r = interest rate and
n = period
Since the interst is compounded quartly,
period = (5×4)-3 = 17
A = 7100(1+2.8/100)∧17
= 7100×1.028∧17
= 11,353.80
The money she will end up earning in interest on the cd = $11,353.80
Answer:
Pr(X-Y ≤ 44.2) = 0.5593
Step-by-step explanation:
for a certain breed of terrier
Mean(μ) = 72cm
Standard deviation (σ) = 10cm
n = 64
For a certain breed of poodle
Mean(μ) = 28cm
Standard deviation (σ) = 5cm
n = 100
Let X be the random variable for the height of a certain breed of terrier
Let Y be the random variable for the height of a certain breed of poodle
μx - μy = 72 -28
= 44
σx - σy = √(σx^2/nx + σy^2/ny)
= √10^2/64 + 5^2/100
= √100/64 + 25/100
= √ 1.8125
= 1.346
Using normal distribution,
Z= (X-Y- μx-y) / σx-y
Z= (44.2 - 44) / 1.346
Z= 0.2/1.346
Z= 0.1486
From the Z table, Z = 0.149 = 0.0593
Φ(z) = 0 0593
The probability that the difference of the observed sample mean is at most 44.3 is Pr(Z ≤ 44.2)
Recall that if Z is positive,
Pr(Z≤a) = 0.5 + Φ(z)
Pr(Z ≤ 44.2) = 0.5 + 0.0593
= 0.5593
Therefore,
Pr(X-Y ≤ 44.2) = 0.5593
Answer:
3.84% probability that it has a low birth weight
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:
If we randomly select a baby, what is the probability that it has a low birth weight?
This is the pvalue of Z when X = 2500. So
has a pvalue of 0.0384
3.84% probability that it has a low birth weight
