Answer:
UV=29
Step-by-step explanation:
In right triangles AQB and AVB,
∠AQB = ∠AVB ...(i) {Right angles}
∠QBA = ∠VBA ...(ii) {Given that they are equal}
We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle
∠AQB+∠QBA+∠BAQ=180 ...(iii)
∠AVB+∠VBA+∠BAV=180 ...(iv)
using (iii) and (iv)
∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV
∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))
∠BAQ=∠BAV...(v)
Now consider triangles AQB and AVB;
∠BAQ=∠BAV {from (v)}
∠QBA = ∠VBA {from (ii)}
AB=AB {common side}
So using ASA, triangles AQB and AVB are congruent.
We know that corresponding sides of congruent triangles are equal.
Hence
AQ=AV
5x+9=7x+1
9-1=7x-5x
8=2x
divide both sides by 2
4=x
Now plug value of x=4 into UV=7x+1
UV=7*4+1=28+1=29
<u>Hence UV=29 is final answer.</u>
I think not because a square has four sides and a triangle has three.
English...because english subject is a part of our language worldwide and that can communicate all over the world...its just my opinion
Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 0°C and a standard deviation of 1.00°C.
This means that 
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69
has a p-value of 0.0455
X = -2.23
has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch:
