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3 years ago

Is this right?? Picture

Mathematics

2 answers:

mixas84 [53]3 years ago

5 0

Ah, not quite. The formula for this, I believe, is 2(wl+hl+hw)
Meaning your equation would look like: 2(5*12+8*12+8*5)
Basically, your final answer is 392 for surface area, I believe.

Aleonysh [2.5K]3 years ago

3 0

Well first you would do 2(lw+hl+wh). Which is 2(12(5)+12(8)+8(5). If you do the math that would be 2(60+96+40). Then add to get 186. Multiply by 2 to get 372. I hope this wasn't confusing.

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ASAP someone please help. Given the following angle measurements for a triangle, find the remaining angle.

Brilliant_brown [7]

Answer:

Step-by-step explanation:

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3 years ago

Round to the nearest whole number of degrees. Enter only a whole number.

quester [9]

Answer:

22 degrees

Step-by-step explanation:

using inverse tangent

2.1/5.3 = 0.396226415

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ANSWER = 21.6147

8 0

3 years ago

Read 2 more answers

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

Find the product.<br> enter your answer in simplest form.<br><br> 6x2/3 = _=_

Ivenika [448]

Answer: 4

Step-by-step explanation:

6 • 2/3 =
12/3 =
4

6 0

3 years ago

A bank account has a beginning of 560.00. After 6 months, the balance in the account has increased to 572.60. What interest rate

MaRussiya [10]

This is tricky because we are given the interest rate for the year but the problem is figureing the interest for 6 months or 1/2 year. We will have to double the difference before solving for the yearly interest.

572.60 - 560.00 = 12.60 interest added for 6 months x 2 = 25.20 for 12 months

(This problem assumes the interest will stay the same the next 6 months)

We need to find what percent 25.20 is to our original balance.

25.20/560 = x / 100

2520 = 560x

x = 4.5 percent interest

Check .045 x 560.00 = 25.20 interest in one year (or 12.60 in 6 months)

4 0

3 years ago