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muminat
3 years ago
13

Please Help!!!!! I have a math test and I am not good at this stuff. and here is a hint: don't use the 10.

Mathematics
1 answer:
Talja [164]3 years ago
5 0
A. 12•8=96, 96/2=48, 48•4=192, 12•12=144, 192+144=336
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What 18% more than 200
djyliett [7]
Answer: 236

Explanation:

100% = 200
118% = x

-> x = (118 * 200) / 100

-> x = 236
3 0
3 years ago
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Evaluate the given numerical expression <br> (5+7 to the 3rd power divided by 7 times 7
-BARSIC- [3]

{(5 + 7)}^{3}  \div 7 \times 7 \\  {12}^{3}  \div 7 \times 7 \\ 1728 \div 7 \times 7 \\ 1728

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3 years ago
What are the powers of three that are in the range 3 through 1000
kirill115 [55]
3^1
3^2
3^3
3^4
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8 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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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
PLEASE HELP!!!!!<br> Combine the like terms to create an equivalent expression. 8k+5k
horsena [70]

Answer:

13k

Step-by-step explanation:

Given: 8k+5k

Finding the equivalent expression.

We know the rule of addition is that adding two positive integers will give positive result or sum.

Now, lets find the equivalent expression of 8k+5k.

∴ 8k+5k= 13k

Hence, the required equivalent expression is 13k.

3 0
3 years ago
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