Throughout this entire problem, I'm going to assume that point S is on line RT.
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Part A)
Because S is on line RT, this means that RS + ST = RT by the segment addition postulate. Make the proper substitutions (aka replacements) and solve for y
RS + ST = RT
(8y+4) + (4y+8) = 15y-9
8y+4 + 4y+8 = 15y-9
(8y+4y)+(4+8) = 15y-9
12y+12 = 15y-9
12y+12-12y = 15y-9-12y
12 = 3y-9
12+9 = 3y-9+9
21 = 3y
3y = 21
3y/3 = 21/3
y = 7
The value of y is y = 7
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Part B)
Since y = 7 (from part A above), we can find the following
RS = 8*y+4
RS = 8*7+4
RS = 56+4
RS = 60
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ST = 4*y+8
ST = 4*7+8
ST = 28+8
ST = 36
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RT = 15*y-9
RT = 15*7-9
RT = 105-9
RT = 96
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Note how RS+ST = 60+36 = 96 and how this is equal to RT = 96 above, so that helps confirm the right answer. It helps confirm that RS+ST = RT.
Answer:
Step-by-step explanation:
Let us assume that the length of fishes in the tank is normally distributed. Thus, x is the random variable representing the length of fishes in the tank. Since the population mean and population standard deviation are known, we would apply the formula,
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = standard deviation
number of samples
From the information given,
µ = 15
σ = 4
n = 31
We want to find the probability that x is between (15 - 0.5) inches and (15 + 0.5) inches. The probability is expressed as
P(14.5 ≤ x ≤ 15.5)
For x = 14.5
z = (14.5 - 15)/(4/√31) = - 0.7
Looking at the normal distribution table, the probability corresponding to the z score is 0.2420
For x = 15.5
z = (15.5 - 15)/(4/√31) = 0.7
Looking at the normal distribution table, the probability corresponding to the z score is 0.7580
Therefore,
P(14.5 ≤ x ≤ 15.5) = 0.7580 - 0.2420 = 0.5160
50% of 20 is 10, 75% of 10 is 7.5. round up and you get 80%.
hope this helps you! :-)
<u>#7</u>
1. Find the volume of the pyramid (l×w×h/3); you should get 64cm³
2. Find the volume of the cube (l×w×h); you should get 512cm³
3. Add the two products; the glass paperweight's volume is 576cm³
<u>#8</u>
1. Find the surface area (SA) of the cube (area of one square×5); it is 320cm²
2. Find the SA of the pyramid (exclude the square base); it is 80cm² ((slant height×(length/2))×4)
3. Add the surface areas; the sum should be 400cm²