Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean 
and standard deviation 
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:
Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:
Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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Circumference is equal to diameter multiplied by pi. C=d X pi. To find the radius we need to rearrange this equation. We know what pi and the circumference are, so to find diameter we do circumference divided by pi. D=c/pi. 942/pi = 299.85. This gives us the diameter, for the radius simply divide by 2 which is 149.92 :)
From the given figure ,
RECA is a quadrilateral
RC divides it into two parts
From the triangles , ∆REC and ∆RAC
RE = RA (Given)
angle CRE = angle CRA (Given)
RC = RC (Common side)
Therefore, ∆REC is Congruent to ∆RAC
∆REC =~ ∆RAC by SAS Property
⇛CE = CA (Congruent parts in a congruent triangles)
Hence , Proved
<em>Additional</em><em> comment</em><em>:</em><em>-</em>
SAS property:-
"The two sides and included angle of one triangle are equal to the two sides and included angle then the two triangles are Congruent and this property is called SAS Property (Side -Angle-Side)
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Consider the proof. Given: Segment AB is parallel to line DE. Prove: AD/DC = BE/EC What is the missing statement in Step 5? A.) AC = BC B.) AC/DC = BC/EC C.) AD...
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Sides are always equal. Pretty sure
