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V125BC [204]
2 years ago
15

In scientific notation, how do you know the number is larger than one?

Mathematics
1 answer:
Anarel [89]2 years ago
7 0
<h3>You can compare two numbers written in scientific notation by looking at their powers of 10. The number with the greater power of 10 will be the greater number. If two numbers have the same power of 10, then compare the decimal numbers to determine the greater number.</h3>
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A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.
noname [10]

Answer:

Width = 4ft

Height = 4ft

Length = 8ft

Step-by-step explanation:

Given

Volume = 128ft^3

L = 2W

Base\ Cost = \$9/ft^2

Sides\ Cost = \$6/ft^2

Required

The dimension that minimizes the cost

The volume is:

Volume = LWH

This gives:

128 = LWH

Substitute L = 2W

128 = 2W * WH

128 = 2W^2H

Make H the subject

H = \frac{128}{2W^2}

H = \frac{64}{W^2}

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

A = LW + 2(WH + LH)

The cost is:

Cost = 9 * LW + 6 * 2(WH + LH)

Cost = 9 * LW + 12(WH + LH)

Cost = 9 * LW + 12H(W + L)

Substitute: H = \frac{64}{W^2} and L = 2W

Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)

Cost =18W^2 +  \frac{768}{W^2}*3W

Cost =18W^2 +  \frac{2304}{W}

To minimize the cost, we differentiate

C' =2*18W +  -1 * 2304W^{-2}

Then set to 0

2*18W +  -1 * 2304W^{-2} =0

36W - 2304W^{-2} =0

Rewrite as:

36W = 2304W^{-2}

Divide both sides by W

36 = 2304W^{-3}

Rewrite as:

36 = \frac{2304}{W^3}

Solve for W^3

W^3 = \frac{2304}{36}

W^3 = 64

Take cube roots

W = 4

Recall that:

L = 2W

L = 2 * 4

L = 8

H = \frac{64}{W^2}

H = \frac{64}{4^2}

H = \frac{64}{16}

H = 4

Hence, the dimension that minimizes the cost is:

Width = 4ft

Height = 4ft

Length = 8ft

8 0
2 years ago
I need help with this math question (Img included)
Mandarinka [93]
Answer: C. 5
Because (x^(2/3))(x^(3/3))=x^(5/3)
5 0
2 years ago
In math class, you have the following scores: 77, 82, and 73. If the
ziro4ka [17]

Answer: 88

Step-by-step explanation: 77+82+73= 232. So you need an 80 for average. So what you have to do is 80x4 and that equals to 320. last step is to subtract 320-232 and you get 88. 77+82+73+88=320.

6 0
3 years ago
Read 2 more answers
Suppose that the demand equation for a certain item is 1p+1x2−160=0. evaluate the elasticity at 65:
bekas [8.4K]
Price elasticity is defined as \frac{\Delta(quantity)}{\Delta(price)}.

Here, the question has omitted to define variables, so we will ASSUME
p=price
x=variable,
and we're given
p+x^2-160=0

We calculate
\delta{x}/\delta{p} by implicit differentiation with respect to p:
1+2x (dx/dp)=0
=>
E(x)=(dx/dp)=-1/(2x)

Therefore the price elasticity at x=65 is
E(65)=1/(2*65) =-1/130 (approximately -0.00769)

4 0
2 years ago
You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r
kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that p = 0.15

Assume you obtain a random sample of 9 individuals from this population:

This means that n = 9

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679

P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0
2 years ago
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