All categories

3 years ago

A square pyramid has a base area of 64 ft2. The pyramid’s total surface area is 224 ft2.

Mathematics

1 answer:

borishaifa [10]3 years ago

5 0

The area of each triangular face is 40 square feet

Solution:

Given that,

Square pyramid has a base area of 64 square feet

The pyramid’s total surface area is 224 square feet

The total surface area of a regular pyramid is the sum of the areas of its lateral faces and its base

Therefore, total surface area of square pyramid is given as,

Total surface area = base area + 4(Area of traingular face)

Let "x" be the area of each traingular face

$224 = 64 + 4(x)\\\\224 - 64 = 4x\\\\4x = 160\\\\x = 40$

Thus area of each triangular face is 40 square feet

You might be interested in

Delta makes 12-volt car batteries. These batteries are known to be normally

blondinia [14]

Answer:

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

Step-by-step explanation:

Step(i):-

Given that the sample size n =12 -volt car batteries

Let 'X' be a Random variable in a normal distribution

Given that mean of the normal distribution = 45 months

Given that the Standard deviation of the normal distribution = 8months

Step(ii):-

Let X₁ = 3 years = 12 × 3 = 36 months

$Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{36-45}{8} = -1.125$

Let X₂ = 4 years = 12 × 4 = 48 months

$Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{48-45}{8} = 0.375$

Step(iii):-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = P(-1.125≤Z≤0.375)

= P(Z≤0.375) - P(Z≤-1.125)

= 0.5 +A(0.375) - (0.5-A(1.125)

= 0.5 + 0.1480 - (0.5 -0.3708)

= 0.1480 + 0.3708

= 0.5188

Final answer:-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

5 0

2 years ago

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$