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gogolik [260]
3 years ago
8

Colton has n nickels and d dimes. He has no more than $2 worth of coins altogether. Write this situation as an inequality.

Mathematics
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

Step-by-step explanation:

Represent the number of nickels and dimes by n and d.

The value of these nickels and dimes will be $0.05n + $0.10d ≤ $2.00.

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(look at image) can someone explain to me how to do number 1
Rufina [12.5K]

Answer:

so it would be y= -4x + 3

Step-by-step explanation:

its asking to write the equation in slope-intercept form so y=mx+b m in this case is -4 and b whihc is the y-intercept is 3 so you just write the equation y= -4x + 3

5 0
2 years ago
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Simplify: 5(3x -4) - 6( 4 - 3x)
VikaD [51]

Answer:

33x-44

Step-by-step explanation:

15x-20-24+18x

33x-44

5 0
3 years ago
Assume that there is a 4​% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit
kap26 [50]

Answer:

a) 99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b) 99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So 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.

4​% rate of disk drive failure in a year.

This means that 96% work correctly, p = 0.96

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 2

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984

99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b. If copies of all your computer data are stored on four independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 4

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744

99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

7 0
3 years ago
What is h a=1/2bh what is h
Ivahew [28]
If it is a triangle h is height and b is base.
8 0
3 years ago
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What is the area of the sector with a central angle of 49° and a radius of 11 cm? Use 3.14 for pi and round your final answer to
olga_2 [115]

Answer:

\boxed{\text{51.71 cm}^{2}}

Step-by-step explanation:

If the angle θ is in radians, the formula for the area (A) of a sector of a circle is

A = ½r²θ

If θ is in degrees

A = \dfrac{1}{2}r^{2}\theta \times \dfrac{\pi \text{ rad}}{180^{\circ}}= \pi r^{2}\times\dfrac{\theta }{360}

Data:

θ = 49°

r = 11 cm

Calculation:

\begin{array}{rcl}A& = &3.14\times 11^{2}\times\dfrac{49}{360}\\\\ & = &3.14 \times 121 \times 0.1361\\ & = & \mathbf{51.71}\\\end{array}\\\text{The area of the sector is }\boxed{\textbf{51.71 cm}^{2}}

4 0
2 years ago
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