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AnnZ [28]
3 years ago
5

Please help me with this!

Mathematics
1 answer:
MrRa [10]3 years ago
5 0

Answer:

The volume of the triangular pyramid is (1/3) of the volume of the triangular prism

Step-by-step explanation:

Volume formulas:

Triangular Pyramid Volume = T = (1/3)*h * (base area)

Triangular Prism Volume = P = h * (base area)

P = 3 * T

and T = (1/3)* P

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Using the graph as your guide, complete the following statement.
Alex787 [66]

Because the parabola intercepts the x-axis only once, we conclude that the discriminant is 0.

<h3>What can we say about the discriminant?</h3>

For a quadratic equation:

y = a*x^2 + b*x + c

The discriminant is:

D = b^2 - 4ac

  • If D = 0, there is only one real zero.
  • If D > 0, there are two real zeros.
  • If D < 0, there are two complex zeros.

In the graph we can see that the parabola intercepts the x-axis in its vertex, then the parabola has only one real zero, then we conclude that the discriminant is equal to zero.

If you want to learn more about quadratic equations:

brainly.com/question/776122

#SPJ1

8 0
2 years ago
A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 102
cupoosta [38]

Answer:

The upper endpoint of the 99% confidence interval for population proportion is 0.13.

Step-by-step explanation:

The confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

<u>Given:</u>

<em>n</em> = 1000

\hat p = 0.102

z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58

*Use the standard normal table for the critical value.

Compute the 99% confidence interval for population proportion as follows:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.102\pm 2.58\times\sqrt{\frac{0.102(1-0.102)}{1000}}\\=0.102\pm0.0248\\=(0.0772, 0.1268)\\\approx (0.08, 0.13)

Thus, the upper limit of the 99% confidence interval for population proportion is 0.13.

6 0
3 years ago
The number of laughs (denoted by L) can be defined as a function of the number of jokes (denoted by J), the amount of knowledge
MrMuchimi

Answer:

(A) (Jokes * Knowledge^2) / Familiarity

Step-by-step explanation:

The number of laughs (denoted by L) is a function of

  • The number of jokes (denoted by J),
  • The amount of knowledge about the joke material (denoted by K).

Given L=\frac{J\cdot K^2}{F}

The appropriate measurement unit for number of laughs will be:

(Jokes * Knowledge^2) / Familiarity

6 0
3 years ago
The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass
Shkiper50 [21]

Half of the starting amount remains after 1 half-life. So after 6 half-lives, the starting amount is halved 6 times, meaning that a 195 g sample would decay down to

(195 g)/2^6 = 195/64 g ≈ 3 g

5 0
3 years ago
What is the equivalent for a 6.5 x * x + 5 .5 x + 1
nadezda [96]

Answer:

6.5x^2 + 5.5x + 1

Step-by-step explanation:

6.5 x * x + 5 .5 x + 1 can be re-written as 6.5x^2 + 5.5x + 1, or as:

65x^2 + 55 + 10

-------------------------

          10

8 0
3 years ago
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