Answer: 228 students
Step-by-step explanation:
Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test reults
µ = mean score
σ = standard deviation
From the information given,
µ = 1700 points
σ = 75 points
We want to find the probability of students expected to score above 1850 points. It is expressed as
P(x > 1850) = 1 - P(x ≤ 1850)
For x = 1850,
z = (1850 - 1700)/75 = 150/75 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.97725
P(x > 1850) = 1 - 0.97725 = 0.02275
If 10,000 students take the exam, then the number of students you would expect to score above 1850 points is
0.02275 × 10000 = 228 students
Answer:
yes because
Step-by-step explanation:
this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for are ‘compatible’ with the given angle-side pair ( ;a) = (30 ;3) in that both choices for can t in a triangle with and both have a sine of 2 3. The only other given piece of information is that c= 4 units.
Answer:
V = 34,13*π cubic units
Step-by-step explanation: See Annex
We find the common points of the two curves, solving the system of equations:
y² = 2*x x = 2*y ⇒ y = x/2
(x/2)² = 2*x
x²/4 = 2*x
x = 2*4 x = 8 and y = 8/2 y = 4
Then point P ( 8 ; 4 )
The other point Q is Q ( 0; 0)
From these two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.
Now with the help of geogebra we have: In the annex segment ABCD is dy then
V = π *∫₀⁴ (R² - r² ) *dy = π *∫₀⁴ (2*y)² - (y²/2)² dy = π * ∫₀⁴ [(4y²) - y⁴/4 ] dy
V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴
V = π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )
V = π * [256/3 - 51,20]
V = 34,13*π cubic units
Answer:
$61.60
Step-by-step explanation:
40.00 X 1.40 = $61.60
