Answer: 1. False.
2.True.
3.True.
4.False.
5.True.
6.False.
Explanation:
1.∠XFG is lying between the one side and the line extended to the next side of the triangle, therefore it is an exterior angle.
2.∠EFG is lying between the two sides of the ΔEFG, therefore it is an interior angle of the triangle.
3.∠FEZ is lying between the one side and the line extended to the other side of ΔEFG.
4.∠YGE is a straight line it is between the one side and the line extended to the same side ∴ it cannot be an exterior angle of the triangle.
5.∠EGF and FGY lying on the same line YZ ,∴they are linear pair i.e. they are supplementary angles.
6.∠FEG and ∠FGE are both the interior angles of a triangle and we know that for supplementary angles they should lie on the same line.Therefore ∠FEG and ∠FGE are not supplementary angles.
We check first the numerical coefficients of both sides of the equation if they match if we perform the operation.
(6)(4) = 24
Then, the variables. For multiplication with the same variables, the exponents are added. In the given above,
n + 2 = 6
The value of n should be 4.
Hi there! Osvoldo did not meet his goal.
(In my answering I suppose you mean 30%, 220 grams of carbohydrates and 55 grams of whole grains, if this is incorrect, please let me know)
Today Osvoldo ate 220 grams of carbohydrates.
10 % of 220 is 22 (divide by 10), and therefore 30 % of 220 is 22 * 3 = 66.
If at least 66 grams of Osvoldo's total consumption consisted of whole grains, he would have met his goal. However, he only ate 55 grams of whole grains (which is less than 66), and therefore he did not meet his goal.
Answer:
This statement can be made with a level of confidence of 97.72%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8.1 mm
Standard Deviation, σ = 0.5 mm
Sample size, n = 100
We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling:

P(mean thickness is less than 8.2 mm)
P(x < 8.2)
Calculation the value from standard normal z table, we have,

This statement can be made with a level of confidence of 97.72%.