No nickles = 2x and that of dimes = 7x so 2x = 350 get x and the no of dimes
so like 2x = 350 divide by 2 on both sides so x = 175 ... and no of dimes = 7*x = 7*175 = 1225
Hope it helps :)
Answer: Choice C) {6, 9, 12}
=======================================================
Explanation:
Something like choice A can't form a triangle because the first two sides add to 6+9 = 15, which is <u>not</u> longer than the third side 15. We need to have x+y > z to be true. The same goes for choice B as well because 3+3 = 6 is too small compared to the third side 7. Choice D is similar to choice A.
In short, we can rule out choices A, B, and D.
The only thing left is choice C. Picking any two sides leads to having that sum be larger than the third side
- 6+9 = 15 which is larger than 12
- 9+15 = 24 which is larger than 6
- 6+15 = 21 which is larger than 9
So the conditions of the triangle inequality theorem work out here, and we have a triangle.
We know this angle is a 90° or a right angle. So another words we know adding these 2 expressions equals to 90°.
(4x + 3) + (x - 8) = 90°
Remove the parentheses and combine the like-terms:
5x - 5 = 90°
Add 5 on both sides:
5x = 95°
Divide 5 on both sides
x = 19
You answer is x = 19.
Answer:
21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.
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:
Calculate the probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.
This is 1 subtracted by the pvalue of Z when X = 9.08. So
has a pvalue of 0.7823
1 - 0.7823 = 0.2177
21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.
