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3 years ago

What is the volume of the triangular prism below? V = ____ ft3

Mathematics

1 answer:

Lelechka [254]3 years ago

8 0

Answer:

4.5

Step-by-step explanation:

triangular prism volume = $\frac{1}{4}$ h $\sqrt{-a^{2}+2(ab)^{2} +2(ac)^{2}﹣b^{4}+2(bc)^{2}﹣c^{4} }$

a = base side = 1.4 ft (after converted from inches to feet)

b = base side= 2 ft

c = base side = 2.5 ft

h = height = 3 ft

plug all the number into the formula

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Please help look at picture <br><br><br> round to the nearest tenth

Salsk061 [2.6K]

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3 years ago

A four-sided die is loaded in a way that the probability of a face is proportional to the number of dots (there are 1, 2, 3, and

alisha [4.7K]

Answer:

a) The probabilities are:

Pr(1) = 1/4

Pr(2) = 1/4

Pr(3) = 1/4

Pr(4) = 1/4

b) Pr(sum of 6 or more on two throw) = 6/16 = 3/8.

c) If the die was fair, it should have six (6) numbers => (1,2,3,4,5,6) and the sample space from two throw will be $6^{2}$ = 36.

Hence, Pr(sum of 6 or more on two throw) = 26/36 = 13/18.

Step-by-step explanation:

a) Since the probability of a face is proportional to the number of dots (there are 1, 2, 3, and 4 dots on the respective sides), then the occurrence of any outcome in a single throw are equally likely. Hence, there is equal probability of occurrence.

By probability, we say

Pr(x) = number of favorable event/ total number of possible events

Therefore, the total number = 4, and each can occur at equal probability. This is why the answer are:

Pr(1) = 1/4

Pr(2) = 1/4

Pr(3) = 1/4

Pr(4) = 1/4

b) The sample space when the die is thrown twice = $4^{2}$ = 16 possible outcome.

Among the outcomes, those that will have 6 or more when sum are just six (6). They are:

(2,4), (3,3), (3,4), (4,2), (4,3), and (4,4)

Thus, Pr(sum of 6 or more on two throw) = 6/16 = 3/8.

Just for understanding:

Other outcome from the sample space are:

(1,1), (1,2), (1,3), (1,4)

(2,1), (2,2), (2,3), ___

(3,1), (3,2), ___, ___

(4,1), ___, ___, ___

If you observe, the dash are the ones we have listed above!

c) When the die is fair, this is what happen.

Just for understanding:

Other outcome from the sample space when the die is fair are:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

The bold outcome are values when sum are 6 or more. And there are 26 of them.

5 0

3 years ago

WILL GIVR YOU BRAINLEST PLEASE HELP ASAP!

netineya [11]

Sorry the question you typed is kinda messy. But I see you have typed “23* 26” I’m gonna assume you’re asking which postulate or theorem is angle 2 and angle 3, and angle 2 and angle 6. So angle 2 and 3 are vertical angles. Angle 2 and 6 are corresponding angles. If that wasn’t what you wanted, please let me know in the comments so I can understand your problem clearly and solve it.

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3 years ago

Pls help this is due tomorrow-

Vladimir79 [104]

Answer:

Step-by-step explanation:

1- $1 \frac{34}{55}$

2- $1 \frac{1}{18}$

3- $\frac{67}{72}$

4- $\frac{19}{24}$

5- $1 \frac{7}{33}$

6- $\frac{19}{20}$

7- $1 \frac{1}{3}$

8- $\frac{5}{7}$

9- $\frac{5}{6}$

10- $\frac{7}{12}$

4 0

3 years ago

HELP GEOMETRY PLEASE

Alexandra [31]

For each of the problems, twice the angle formed by the chords is equal to half the sum of the angles of the arcs.

So, for the first problem, we have $94 \cdot 2 = x+103$ , so $x = 188 - 103 = 85$ .

For the second, $102 \cdot 2 = x + 118$ , so $x = 204 - 118 = 86$ .

For the last problem, $106 \cdot 2 = x + 88$ , so $x = 212 - 88 = 124$ .

Feel free to comment below if you have any questions!

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3 years ago