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Elza [17]
3 years ago
12

Need help asap please no scammers!!!!!!!!!!!!!!!

Mathematics
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

111

Step-by-step explanation:

u got scammed

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Order from least to greatest. 2.54, 20/ 9 , 32/ 15 , 2.62
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32/15, 20/9, 2.54, 2.62
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I need to know the lengths of MN, NP, and PM
andrezito [222]

Answer:

MN = 12 units

NP = 16 units

PM = 20 units

Step-by-step explanation:

You can find MN and NP by counting on the graph. To find PM you will need to use the formula a^2 + b^2 = c^2 ( ^ stands for exponent)

MN and NP will be a and b.

Example: 12^2 + 16^2 = c^2

Simplify.

Example:

144 + 256 = c^2

144 + 256 = 400

Then find the square root of 400 which is 20.

To check your work put all of the sides into the formula and it should be true.

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3 years ago
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What is the equivalent fraction for 3,2,6,4 whole number?
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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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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:

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Thus, the sample selected must be of size 240.

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3 years ago
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3 years ago
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