I am really stuck at these 4 questions. Pls help:

a) Two spheres of the same density have a ratio of 9 to 16 inches in surface area, If the small sphere weighs 15 kg what does the sphere weighs. Answer to the nearest 10th of KG.
b) The ratio of the volume of two cubes is 216 to 64. What is the ratio of their surface areas?
c) Determine a 90% confidence interval for µ, δ=5, x¯= 70 and n=82. Answer interval rounded to two decimal places.
d) How many people would need to be surveyed on health care to estimate the mean of an entire population to within 2 units if the std deviation was 10 and we wanted a 92% confidence level?

a) To find the weight of the larger sphere, we can set up a proportion using the surface area ratio.

Let x be the weight of the larger sphere:
(9/16) = (15/x)

Cross-multiplying:
9x = 16 * 15
9x = 240

Dividing both sides by 9:
x = 240/9
x ≈ 26.7 kg

Therefore, the weight of the larger sphere is approximately 26.7 kg.

b) The ratio of the surface areas of two cubes is equal to the square of the ratio of their side lengths.

Let a and b be the side lengths of the cubes:
Surface Area Ratio = (a^2)/(b^2)

Given that the volume ratio is 216/64, we can find the side length ratio by taking the cube root of the volume ratio:
Side Length Ratio = cube root of (216/64) = 6/4 = 3/2

Substituting the side length ratio into the surface area ratio equation:
Surface Area Ratio = (3/2)^2 = 9/4

Therefore, the ratio of their surface areas is 9 to 4.

c) To determine the confidence interval for µ (population mean), we can use the formula:

Confidence Interval = x¯ ± (Z * (δ/√n))

In this case, δ (population standard deviation) is given as 5, x¯ (sample mean) is given as 70, and n (sample size) is given as 82.

First, we need to find the Z score corresponding to the desired confidence level of 90%. Using a Z-table or calculator, we find that the Z score for a 90% confidence level is approximately 1.645.

Now, we can calculate the confidence interval:

Confidence Interval = 70 ± (1.645 * (5/√82))

Calculating the values in the formula:
Confidence Interval = 70 ± (1.645 * 0.55115)

Calculating the lower and upper bounds of the interval:
Lower Bound = 70 - (1.645 * 0.55115)
Upper Bound = 70 + (1.645 * 0.55115)

Rounding the bounds to two decimal places:
Lower Bound ≈ 69.10
Upper Bound ≈ 70.90

Therefore, the 90% confidence interval for µ is approximately (69.10, 70.90).

d) To estimate the mean of an entire population within 2 units with a 92% confidence level, we can use the formula:

Sample Size (n) = (Z^2 * σ^2) / (E^2)

In this case, the standard deviation (σ) is given as 10, the desired margin of error (E) is 2, and the Z score corresponding to a 92% confidence level can be found using a Z-table or calculator (approximately 1.751).

Now, we can calculate the sample size:

n = (1.751^2 * 10^2) / (2^2)
n = (3.064501 * 100) / 4
n ≈ 76.61

Rounding up to the nearest whole number, we need at least 77 people to be surveyed on health care to estimate the population mean within 2 units with a 92% confidence level.

a) To find the weight of the larger sphere, we first need to find the ratio of their volumes. The ratio of their surface areas is given as 9:16. We know that the surface area of a sphere is proportional to the square of its radius (or diameter).

Let the radius of the small sphere be r.
The surface area of the small sphere = 4πr^2

Now, let the radius of the larger sphere be R.
The surface area of the large sphere = 4πR^2

According to the given ratio, (4πR^2) / (4πr^2) = 9/16

Simplifying, we have R^2 / r^2 = 9/16

Taking the square root of both sides, we get R/r = 3/4

Since the densities of the two spheres are the same, we can say that the ratio of their volumes is also (R^3) / (r^3) = (3/4)^3

Now, we are given that the weight of the small sphere is 15 kg. Since weight is proportional to volume, we can set up the following equation:

(Weight of larger sphere) / (Weight of small sphere) = (Volume of larger sphere) / (Volume of small sphere)

Let the weight of the larger sphere be W.
(Weight of larger sphere) / 15 = [(R^3) / (r^3)]

Substituting the value of (R^3) / (r^3) from the previous step, we have:
W / 15 = [(3/4)^3]

Simplifying, W = 15 * [(3/4)^3]

Calculating this value will give you the weight of the larger sphere, rounded to the nearest 10th of a kilogram.

b) The ratio of the volumes of the two cubes is given as 216:64. Since the volume of a cube is proportional to the cube of its side length, we can set up the following equation:

(Volume of larger cube) / (Volume of smaller cube) = (Side length of larger cube)³ / (Side length of smaller cube)³

Let the side length of the larger cube be L.
The ratio of their volumes is (L^3) / (L^3/3) = 216/64

Simplifying, we have L^3 / (L^3/3) = 216/64

Multiply both sides by L^3/3 to get: L^3 = (L^3/3) * 216/64

Simplifying further, L^3 = L^3 * (3/1) * (27/8)

This equation is not possible, as it implies that 1 = 3 * 27/8, which is not true.

Therefore, there is an error in the question, as the given ratio of volumes is not accurate.

c) To determine a 90% confidence interval for µ (population mean), we can use the formula:

Confidence Interval = x¯ ± (Z * (δ / √n))

Where:
- x¯ is the sample mean
- Z is the z-score corresponding to the desired confidence level
- δ is the population standard deviation
- n is the sample size

In this case, we are given:
- x¯ = 70
- δ = 5
- n = 82

To find the z-score corresponding to a 90% confidence level, we can refer to a standard normal distribution table or use a calculator. The z-score for a 90% confidence level is approximately 1.645.

Substituting the values in the formula, the confidence interval is:
70 ± (1.645 * (5 / √82))

Calculating this will give you the confidence interval, rounded to two decimal places.

d) To estimate the mean of an entire population within a certain margin of error, we can use the formula:

n = (Z² * (σ²)) / (E²)

Where:
- n is the required sample size
- Z is the z-score corresponding to the desired confidence level
- σ is the population standard deviation
- E is the desired margin of error

In this case, we want a 92% confidence level and a margin of error of 2 units. The z-score for a 92% confidence level is approximately 1.751.

Substituting the values in the formula, we have:
n = (1.751² * (10²)) / (2²)

Calculating this will give you the required sample size.