The Graduate Management Admission Test (GMAT) is a standardized exam used by many universities as part of the assessment for adm
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Answer:
Step-by-step explanation:
Suppose pieces of chocolate pie served at a restaurant average 350 calories with a standard deviation of 20 calories. Calories have a Normal distribution due to inexact cutting of the pies. Which graph represents the proportion of pieces of pie that have more than 375 calories.
Answer: The z score is used to determine how many standard deviations that the raw score is above or below the mean. If the z score is positive then the raw score is above the mean and for a negative z score means the raw score is below the mean. The z score is given as:
Given that: μ = 350 calories, σ = 20 calories, x > 375
From the graph, The shaded area represents the proportion of pieces of pie that have more than 375 calories
From the normal distribution table, P(x > 375) = P(z > 1.25) = 1 - P(z < 1.25)
= 1 - 0.8944 = 0.1056 = 10.56%
<u>Solution- </u>
A researcher placed a petri dish with 32,000 bacterial cells. One hour after being placed in the vacuum chamber, the number of cells in the petri dish had halved. Another hour later, the number of cells had again halved.
This can be represented as exponential decreasing function,
Where,
- a = starting amount = 32000
- r = rate = 50% = 0.5 as the sample becomes halved in each hour
- x = hours
Putting the values,
y-intercept means, where x=0, so
The coordinate of this poin will be (0, 32000)
This means when x=0 or at the starting of the research, the number of bacteria cells was 32000.
After 3 hours, number of bacteria cells will be,
The coordinate of this point will be (3, 4000)
Answer:
a) the length of the wire for the circle =
b)the length of the wire for the square =
c) the smallest possible area = 126.02 in² into two decimal places
Step-by-step explanation:
If one piece of wire for the square is y; and another piece of wire for circle is (60-y).
Then; we can say; let the side of the square be b
so 4(b)=y
b=
Area of the square which is L² can now be said to be;
On the otherhand; let the radius (r) of the circle be;
2πr = 60-y
Area of the circle which is πr² can now be;
Total Area (A);
A =
=
For the smallest possible area;
∴
If we divide through with (2) and each entity move to the opposite side; we have:
By cross multiplying; we have:
2πy = 480 - 8y
collect like terms
(2π + 8) y = 480
which can be reduced to (π + 4)y = 240 by dividing through with 2
∴ since , we can determine for the length of the circle ;
60-y can now be;
=
=
=
=
also, the length of wire for the square (y) ;
The smallest possible area (A) =
= 126.0223095 in²
≅ 126.02 in² ( to two decimal places)