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3 years ago

10. what is the volume of the prism to the nearest whole unit?

Mathematics

2 answers:

Ainat [17]3 years ago

6 0

To calculate the volume of the rectangular prism and the triangular prism you will multiply the area of the base by the height of the prism.

1. V = BH
V = lwH
12 x 8 x 4
V = 384 cubic inches

2. V = BH
V = 1/2bhH
1/2 x 12 x 2 x 18
V = 216 cubic feet

s2008m [1.1K]3 years ago

4 0

Answer:

10. Option 1

11. Option 3

Step-by-step explanation:

Given the dimensions of rectangular prism. we have to calculate the volume of prism.

Length, l=12 inches

Width, w=8 inches

Height, h=4 inches

$Volume=lwh=12\times 8\times 4=384\thinspace in^3$

Option 1 is correct.

Given the dimensions of triangular prism. we have to calculate the volume of prism.

Length, H=18 ft

Base, b=12 ft

Height, h=2 ft

$V=\frac{1}{2}bhH=\frac{1}{2}\times 12\times 2\times 18=216ft^3$

Option 3 is correct.

You might be interested in

PLEASE HELP!!!!!!!!

Korolek [52]

Answer:

2π - 4√5

(= -2.66 to nearest hundredth).

Step-by-step explanation:

√(2π - 4√5)^2

= 2π - 4√5 (by the definition of a square root)

If you want an approximate value it is

-2.66 to the nearest hundredth

7 0

2 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

3 years ago

Please help i really need an A and i'm horrible at math ):

Dovator [93]

See the attached picture for 2 options for question 1 and 2 depending on how the fractions are written.

Question #3:

17 = -13 - 8x

Add 13 to each side:

30 = -8x

Divide both sides by -8:

X = 30 / -8

X = -3 3/4

4 0

3 years ago

Select all ratios equivalent to 30:6

Korolek [52]

Answer:

60 : 12

90 : 18

120 : 24

150 : 30

3 0

3 years ago

Sam made $210.00 last week. If he worked 35 hours, how much is he paid for one hour of work? Answer: dollars per hour.

timama [110]

Answer:

$6 Dollars Per Hour

Step-by-step explanation:

210/35=6

8 0

3 years ago