Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
C) Between 80 and 120
D) less than 80
E) Between 70 and 100
F) More than 130
We can use ratios, in other words, if one part of the photo is changed x times, how many times does the other part change?
We can set up the proportion as :
width 1 : length 1 = width 2 : length 2
20 : 30 = 5 : x
We do inverse multiplication, or multiply the outer with outer and inner with inner (W1 * L2 and L1 * W2)
20*x = 30*5
20x = 150
Divide all by 10
2x = 15
Divide all by 2 to isolate x
x = 15/2
x = 7.5cm
Answer:
I don't get what you are trying to ask?
Sin (A + B) = sin A cos B + cos A Sin B
<span>Cos (A - B) = cos A cos B + sin A sin B </span>
<span>=> (SinACosB+ CosASinB) (CosACosB +SinASinB) </span>
<span>=>SinACosACos^2B+Sin^2ACosBSinB+Cos^2A... </span>
<span>=>SinACosA(Cos^2B+Sin^2B) +SinBCosB(Sin^2A+Cos^2A) </span>
<span>we know that Sin^2+Cos^2=1 </span>
<span>=>SinACosA(1)+SinBCosB(1) </span>
<span>=SinACosA+SinBCosB </span>
<span>Proved
</span>
