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mario62 [17]
4 years ago
15

Which ordered pair is a solution to the equation?

Mathematics
2 answers:
____ [38]4 years ago
6 0
It’s The very first -2,3
saveliy_v [14]4 years ago
4 0
C is the answer hope I helped
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A digital gaming service offers a subscription plan with a $25 membership fee and charges a $2.50 per game rental fee. How many
erastova [34]

Answer:

They can get at most 10 games

The inequality is g=25+2.50x

Step-by-step explanation:

50=25+2.5x

50-25=25

25=2.5x

25/2.5=10

6 0
3 years ago
Can you answer problem 11?
In-s [12.5K]

Answer:

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7 0
3 years ago
For options A– E choose all of the expressions that are equivalent to 3(4x + 3).
Katen [24]
3(4x+3) = 12x + 9
A= 3(4x+1)
B= 9 + 12x
C= 7x + 6
D= 12x + 9
E= 12x + 9

Answer: D, E

Hope This Helps✨
3 0
3 years ago
A large farm has 75 acres of wheat and 62.5 acres of corn. The farm crew can harvest the the wheat from 12 acres and the corn fr
Leviafan [203]
In a Farm, there are:
=> 75 acres of wheat
=> 62.5 acres of corn
In each day, the farm crew can harvest:
=> 12 acres of wheat
=> 10 acres of corn
Find how many days can the farm crew harvest all of the plants,
=> 75 / 12
=> 6.25, thus the farm crew will take 6.25 days to be able to harvest acres of wheat
=> 62.5 / 10
=> 6.25, thus the farm crew will take 6.25 days also to harvest acres of corn.



4 0
4 years ago
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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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.

5 0
3 years ago
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