Ilene who is not yet 21 years old, is two years older than ida

The cone pictured has a surface area of ______ square centimeters. (Use 3.14 for π.)

Marina86 [1]

Assuming that the units are in cm
Given,

Diameter of the cone = 12 cm

Using the above, the radius of the cone can be determined, which is half the diameter of the cone

∴ Radius of the cone (r) = 12/2 = 6 cm
Height of the cone (h) = 7 cm

Using the above, slant height (s) = √(r² + h²) = √(6² + 7²) = √36 + 49 = √85 = 9.22 cm

π = 3.14

Now using the above values, we can calculate the total surface area of the cone.
Total surface area of the cone (A) = Lateral surface area of the cone (L) + Base surface area of the cone (B)

Lateral surface area of the cone (L) = π r s = 3.14 x 6 x 9.22 = 173.70 cm²
Base surface area of the cone (B) = π r² = 3.14 x 6² = 113.04 cm²

Total surface area of the cone = L + B = 173.70 + 113.04 = 286.74 cm²

6 0

1 year ago

Combine the following expressions . sqrt(9m ^ 2 * n^ 2) + 2sqrt(m ^ 2 * n ^ 2) - 3mn

storchak [24]

Answer:

$2mn$

Step-by-step explanation:

$\sqrt{9m^2*n^2} + 2\sqrt{m^2*n^2} -3mn$

6 0

1 year ago

Oil wells are expensive to drill, and dry wells are a great concern to oil exploration companies. The domestic oil and natural g

Grace [21]

Answer:

(a) H₀: p₁ ≤ p₂ vs. Hₐ: p₁ > p₂.

(b) The point estimate of the proportion of wells drilled in 2005 that were dry is 0.202.

(c) The point estimate of the proportion of wells drilled in 2012 that were dry is 0.111.

(d) The p-value of the test is 0.017.

The wells drilled in 2005 were more likely to be dry than wells drilled in 2012.

Step-by-step explanation:

The data provided for the wells drilled in 2005 and 2012 are as follows:

2005 2012

Wells drilled 119 162

Dry Wells 24 18

(a)

The hypothesis to test whether the wells drilled in 2005 were more likely to be dry than wells drilled in 2012 are defined as follows:

H₀: The wells drilled in 2005 were not more likely to be dry than wells drilled in 2012, i.e. p₁ ≤ p₂.

Hₐ: The wells drilled in 2005 were more likely to be dry than wells drilled in 2012, i.e. p₁ > p₂.

(b)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point-estimate of the population mean μ.

Similarly the point estimate of population proportion p is, $\hat p$ .

It is computed using the formula:

$\hat p=\frac{X}{n}$

Compute the point estimate of the proportion of wells drilled in 2005 that were dry as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{24}{119}=0.202$

Thus, the point estimate of the proportion of wells drilled in 2005 that were dry is 0.202.

(c)

Compute the point estimate of the proportion of wells drilled in 2012 that were dry as follows:

$\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{18}{162}=0.111$

Thus, the point estimate of the proportion of wells drilled in 2012 that were dry is 0.111.

(d)

A z - test for two proportions will be used to perform the test.

The test statistic is defined as:

$z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat P(1-\hat P)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$

Compute the value of standard error as follows:

$SE=\sqrt{\hat P(1-\hat P)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}\\\hat P=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{24+18}{119+162}=0.15\\SE=\sqrt{0.15(1-0.15)(\frac{1}{119}+\frac{1}{162})}\\=0.043$