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koban [17]
4 years ago
15

PLEASE HURRY FIRST ANSWER WILL BE BRAINLIEST can someone translate these word problems to an actual equation?

Mathematics
1 answer:
Reika [66]4 years ago
3 0

Answer:

1). 9 red 6 fancy 2). 50 shirts 20 pants 3). 62 purple 31 red 4). 19 dimes 9 quarters

Step-by-step explanation:

Most are systems of equations. So, they will have more than one equation.

1). x (Red wrapping paper) y (fancy print paper)

x + y = 15

2x + 4y = 42

Solved: x = 9 y = 6

2). x (shirts) y (pants)

x + y = 70

12x + 30y = 1200

Solved: x = 50 y = 20

3). x (purple) y (red)

I ended up dividing 93 by 3 and then multiplying it by 2 for purple and keeping the other third for red. Not quite sure how to fit this into an equation, its just simple math.

Solved: x = 62 y = 31

4). x (dimes) y (quarters)

.10x + .25y = 4.15

x = y - 10

Solved: x = 19 y = 9

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Well technically the check will not go through because she does not have a balance greater than or equal to $55, but if you need an answer it would be $47-$55= $-8. 
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An object falls from a height of 400 feet. Its height, h, in feet is given by the equation h=-16t^2+400 , where t is the time in
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I need help finding the deference from 80-26
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The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

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3 years ago
The sum of 10 and twice a number<br> (an equation)
Gekata [30.6K]

Answer:

10+2x= the sum (anything could be put here)

Step-by-step explanation:

Well sum is automatically addition

Doubleing a number is basically multiplying it by 2

so

10+2x= whatever youw ant to put here

8 0
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