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3 years ago

9

How much greater is the volume of a triangular prism with base area of 15 cm^2 and height of 3.8 cm than the volume of a triangu

lar pyramid with the same height and base area

1 answer:

frosja888 [35]3 years ago

6 0

28.5 is the Answer i thinck i do not now i forgot how to do it in college we do not do that any more

You might be interested in

Find the area and the circumference of a circle with diameter 6 m.

Strike441 [17]

Zxzddsseddzssaaazzsasa

7 0

3 years ago

The two dice are rolled. what is the theoretical chance that they land to form a sum of 11?

Rufina [12.5K]

Answer:

Propability of getting either sum is 2/9

Step-by-step explanation:

Well, we know that there are 36 total arrangements of the two six sided die. So we have a possibility of 3,4 and 4,3. Likewise, we have a possibility of 5,2 and 2,5. We also have a possibility of 1,6 and 6,1. That takes care of the probability That you get two die that add up to 7. Now for 11, you have 5,6 and 6,5. So since you asked for the probability of getting either or, you add all the probabilities, which is 6/36+2/36=8/36=2/9. So the probability of getting either sum is 2/9.

5 0

3 years ago

A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. (a) De

ivanzaharov [21]

Answer:

a) $P(\bar X>81.5)=1-0.933=0.067$

b) $P(\bar X$

c) $P(73.4$

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=77,\sigma=27)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case $\bar X \sim N(77,\frac{27}{\sqrt{81}})$

Part a

We want this probability:

$P(\bar X>81.5)=1-P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >81.5)=1-P(Z$

$P(\bar X>81.5)=1-0.933=0.067$

Part b

We want this probability:

$P(\bar X\leq 69.5)$

If we apply the formula for the z score to our probability we got this:

$P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z$

$P(\bar X\leq 69.5)=0.0062$

Part c

We are interested on this probability

$P(73.4$

If we apply the Z score formula to our probability we got this:

$P(73.4$

$=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}$

And we can find this probability on this way:

$P(-1.2$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-1.2$

3 0

4 years ago

DF and GI are parallel lines.<br><br> Which angles are alternate interior angles?

elena-s [515]

Answer:

your answer is no. d . Because of alternate <

5 0

3 years ago

Num teste de resistência de um para-choques, um automóvel de 2300 kg que circula a 15 m/s, embate num obstáculo. Sabendo que a c

Viktor [21]

Responder:

63.888,89N

Explicación paso a paso:

Fuerza = m (v-u) / t

m es la masa = 2300 kg

v es la velocidad final = 15 m / s

u es la velocidad inicial = 0 m / s

t es el tiempo = 0.54s

Sustituir

F = 2300 (15-0) /0,54

F = 2300 (15) /0,54

F = 34500 / 0,54

F = 63.888,89N

Por lo tanto, la intensidad de la fuerza que no está motorizada es 63,888.89N

7 0

3 years ago