Answer:
0.961
Step-by-step explanation:
To answer this, find the area under the standard normal curve to the left of 130 pounds:
Using the function normalcdf( on a TI calculator, we get:
normalcdf(-1000, 130, 100, 17) = 0.961
Answer:
in steps
Step-by-step explanation:
DE // BC
m∠ADE = m∠ABC and m∠AED = m∠ACB
∴ ΔADE similar to ΔABC
AB/AD = AC/AE
(AD + DB) / AD = (AE + EC) / AE
AD/AD + DB/AD = AE/AE + EC/AE
1 + DB/AD = 1 + EC/AE
DB/AD = EC/AE (AD/DB = AE/EC)
Density = m/v with mass being in grams and volume in mL......
192 grams/250 mL = 0.768 g/mL
Answer:
2236.77 feets
Step-by-step explanation:
From the attached picture, we are to find x, which is the vertical depth of the submarine.
Using trigonometry based on the illustration in the diagram:
Sinθ = opposite / hypotenus
Sinθ = x (depth of submarine) / distance from ship (4000)
Hence,
Sinθ = x / 4000
θ = 34°
Sin34 = x / 4000
0.5591929 * 4000 = x
2236.7716 = x
Hence, depth of submarine = 2236.77 feets
Area = πr²
50 = πr²
r = √50/π
so diameter = 2√50/π
which is = 7.98m to 3sf