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Mila [183]
3 years ago
12

2/(11/16 +3) I know the answer is 12/29. I just don't know how to get there

Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0
11/6+3 is 29/6. 2/(29/6) is 12/29, which is the answer.
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Answer:

True

Step-by-step explanation:

If a quadrilateral (with one set of parallel sides) is an isosceles trapezoid, its legs are congruent.

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What is the value of 3x+3y?
NNADVOKAT [17]

Answer:

3(x+y)

Step-by-step explanation

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7 0
3 years ago
f. If a salesperson receives a commission of 12% of sales, what is the salesperson's commission on $250,000 of sales?​
Gala2k [10]

Answer:

30,000

Step-by-step explanation:

250,000 * .12 = 30,000

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8 0
2 years ago
Mai had 5 candy bars she ate half of one candy bar and dicided to distribute the remaining bars between her two sisters and her
gogolik [260]

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

So there are five candy bars.

Herself and two sisters equals 3 people in total.

This is a graph of 5 candy bars, each line being 1/2.

━ ━

━ ━

━ ━

━ ━

━ ━

If she ate half of one... the graph would become this.

━ ━

━ ━

━ ━

━ ━

━

Now there are 9 halves. You need to split the 9 halves for 3 people. 9 divided by 3 is 3.

Each person gets 3 halves, or 1 and 1 half.

Mai: ━ ━ ━

Sister 1: ━ ━ ━

Sister 2: ━ ━ ━

Altogether that is 9 halves, AKA the number of halves Mai had after she ate 1/2.

The amount Mai ate in the first place: ━

9 halves plus 1 half, equals 10 halves. Each whole has 2 halves. 10 divided by 2 is 5, AKA the number of candy bars she had in the first place.

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

7 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 A 5 mL volume of the suspension is withdrawn.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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