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Mashcka [7]
3 years ago
14

HELP PLZ I'M CONFUSED

Mathematics
2 answers:
SVEN [57.7K]3 years ago
6 0
<span>4 + 3 + 7 ? 7 + 0 +7 

</span>4 + 3 + 7 = 14
7 + 0 +7  = 14
so

4 + 3 + 7  =  7 + 0 +7 
answer
<span>D. =</span>
alukav5142 [94]3 years ago
3 0
.........the answer is D. =
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= { \rm{100\% - 25 \% }} \\ { \rm{ = 75\%}}

• then multiply the percentage decrease with the mass given:

= { \tt{75\% \times 140}} \\  \\  = { \tt{ \frac{75}{100}  \times 140}} \\  \\  = { \underline{ \tt{ \:  \: 105 \: g \: }}}

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Camryn earns $198.60 per week. She works 15 hours each week. How much does Camryn earn per hour? P.S this is a unit rate questio
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Identify 2 items that you can find at a grocery store that hold less than 100 mL.
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3 0
3 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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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