Iris has a family photograph with a width of 8 in and height of 10 in. She wants to enlarg the photograph so that it's width is
2 answers:
Answer:
The enlarge height is 15 in .
Step-by-step explanation:
Let us assume that the width be w .
Let us assume that the height be h .
(As width and height of the photograph are proportional.)
w = kh
Where k is the constant of proportionality .
As given
Iris has a family photograph with a width of 8 in and height of 10 in.
Here w = 8 in , h = 10 in
Put in the equation w = kh .
8 = 10k
She wants to enlarge the photograph so that it's width is 12 in.
Let us assume that the enlarge height be x .
Here w = 12 in , h = x
Put in the equation w = kh .
12 = kx
Compare the value of k .
x = 15 in
Therefore the enlarge height is 15 in .
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Dimensions of the room in cm = 2.54 x 12 by 15 x 2.54 by 2.54 x 8.5 = 30.48 by 38.1 by 21.59
Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm
Given that the density of air at room temperature is
, thus the mass of air in the room = 25,072.21 x 0.00118 = 29.59 g = 0.0296 kg
Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 = 8.88 mg.</span>
Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm
Given that the density of air at room temperature is
Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 = 8.88 mg.</span>
Answer:
For this case we want to check if the true mean for the depth of groves cut into aluminium by a machine is equal to 1.7 (null hypothesis) and the alternative hypothesis would be the complement different from 1.7. And the best system of hypothesis are:
Null hypothesis:
Alternative hypothesis
Alternative hypothesis [tex]\mu \neq 1.7[/tx]
And the best system of hypothesis are:
3. This two-sided test:
H0: μ = 1.7 mm
H1: μ ≠ 1.7 mm
The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50