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MissTica
4 years ago
10

A triangular pyramid has a base with an area of 43 cm squared and lateral faces with bases of 10 cm and heights of 8.6 cm. What

is the surface area of the pyramid?
Mathematics
1 answer:
Jet001 [13]4 years ago
8 0

Answer:

The total surface area of triangular pyramid is 172 cm squared

Step-by-step explanation:

Triangular pyramid:

  • Number of faces 4.
  • Number of vertices of a triangular pyramid is 6.
  • The volume  is \frac13 AH. A= area of the pyramid's base and H= height of the pyramid.
  • The surface area of triangular pyramid B+L. B= area of base, L= area of lateral surface.

Given that, the area of the base is 43 cm squared. Lateral faces with bases of 10 cm and heights 8.6 cm.

The 3 sides of the triangular pyramid is triangle in shape.

The area of triangle is \frac12 \times base \times height.

The lateral surface area of the triangular pyramid is

=3(\frac12 \times base \times height)

=3(\frac12\times 10\times8.6) cm squared

=129 cm squared

The total surface area of triangular pyramid is

=Area of the base + lateral surface area

=(43+129) cm squared

=172  cm squared

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8 0
4 years ago
A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa
Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is  α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion  

p(private) be the proportion of alumni of the private university who attended at least one class reunion  

Hypotheses are:

H_{0}: p(public) = p(private)

H_{a}: p(public) ≠ p(private)

The formula for the test statistic is given as:

z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

  • p1 is the sample proportion of  public university students who attended at least one class reunion  (\frac{808}{1311}=0.616)
  • p2 is the sample proportion of private university students who attended at least one class reunion  (\frac{647}{1038}=0.623)
  • p is the pool proportion of p1 and p2 (\frac{808+647}{1311+1038}=0.619)
  • n1 is the sample size of the alumni from public university (1311)
  • n2 is the sample size of the students from private university (1038)

Then z=\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}} =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.  

6 0
4 years ago
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