Emma enjoys flying kites. her red kite has a maximum altitude of 117.46 meters, and her black kite has a maximum altitude of 362
1 answer:
You might be interested in
Answer:
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given total number of students n(T) = 150
Given 125 of them are fluent in Swahili
Let 'S' be the event of fluent in Swahili language
n(S) = 125
The probability that the fluent in Swahili language
Let 'E' be the event of fluent in English language
n(E) = 135
The probability that the fluent in English language
n(E∩S) = 95
The probability that the fluent in English and Swahili
<u><em>Step(ii):</em></u>-
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = P(S) + P(E) - P(S∩E)
= 0.833+0.9-0.633
= 1.1
<u><em>Final answer:-</em></u>
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Answer:
The cost of the material for the full-sized pad is;
d) 222,750.00
Step-by-step explanation:
The parameters of the steel pad and the scale model are;
The volume of the deck = 18 yd³
The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick
The weight of the sample, m₂ = 75 lbs
The cost of the steel, P = $55/lb
The dimensions of the sample in yards are;
6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0. yards
The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0. yd. = (1/3) yd.³
The density of the sample material, ρ = m₂/V₂
∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³
Therefore, the density of the material of the pad, ρ = 225 lbs/yd³
The mass of the steel pad, m₁ = ρ × V₁
∴ The mass of the steel pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs
The cost of the material for the full-sized steel pad, C = m₁ × P
∴ C = 4,050 lbs × $55/lb = $222,750