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3 years ago

I have to find the surface area of a cereal box. (ohs cereal)

Mathematics

1 answer:

ladessa [460]3 years ago

5 0

Answer:

366 square inches

Step-by-step explanation:

Surface Area Formula:

A= 2( w*l + h*l + h*w)

So you substitute the numbers:

A= 2(3*9 + 13*9 + 13*3)

A= 2( 27+ 117 + 39)

A= 54+234+78

A= 366 square inches

There you go :)

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Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t

jekas [21]

Answer:

(12.1409, 14.0591

Step-by-step explanation:

Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.

Group Group One Group Two

Mean 87.600 74.500

SD 1.000 1.500

SEM 0.316 0.433

N 10 12

The mean of Group One minus Group Two equals 13.100

standard error of difference = 0.556

90% confidence interval of this difference:

$(13.1-1.725*0.556,13.1+1.725*0.556)\\=(12.1409, 14.0591)$

t = 23.5520

df = 20

4 0

3 years ago

What is 212 divided by 8?

hram777 [196]

212 divided by 8 is 26.5. I hope this helps you!

5 0

3 years ago

Read 2 more answers

1 yogurt cost 55p how many can be bought for £5

Alenkinab [10]

If one pound of yogurt cost 55 (cents?) then, Two yogurts cost $1.10
Three yogurts cost $1.65
Four yogurts cost $2.20
Five yogurts cost $2.75
Six yogurts cost $3.30

7 0

3 years ago

The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6

viktelen [127]

Answer:

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

The proportion of contracts involved between five and 10 inclusive bidders is 0.5.

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders No. contracts Relative frequency of contracts

2 6 6/102=0.0588

3 20 20/102=0.1961

4 24 24/102=0.2353

5 18 18/102=0.1765

6 13 13/102=0.1275

7 7 7/102=0.0686

8 5 5/102=0.049

9 6 6/102=0.0588

10 2 2/102=0.0196

11 1 1/102=0.0098

Total 102

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

We have to find proportion of contracts involved between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The proportion of contracts involved between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10= Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

8 0

3 years ago

Simplify. (x6 y3)(−8xy2)(−6x3)

meriva

When simplifying this, we would have to take this little by little in serious steps and not only this, but to also take notice of the use of pemdas and to make sure that we take each step carefully.

We first would (first) simplify the following:

$(((x^6)*(y^3))*(0-(8x*(y^2))))*(0-(2*3x^3)) \\ \\ then \ we \ would \ do . . . . \\ \\ x^7 \ multiplied \ by \ x^3 \ = x^(^7^ +^ 3^) = \ x^1^0 \\ \\ (last \ step).... \\ \\ 2^3 \ multiplied \ by \ 2^1 = 2^(^3^ +^ 1^) = 2^4 \\ \\ \Longrightarrow \ (your \ answer) \\ \\ \boxed{\bf{ (2^4*3x^1^0y^5)}}$

7 0

3 years ago