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Zepler [3.9K]
3 years ago
7

The mean of a set of six numbers is 4.5. If the mean of two of the numbers is 5.5, find the mean of the remaining four numbers

Mathematics
1 answer:
lilavasa [31]3 years ago
3 0
Ummm this could take me time but I can answer it
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The difference between two numbers is 3.50. The lesser number is 1.65.
deff fn [24]

Answer:

5.15

Step-by-step explanation:

you add 3.50 and 1.65 to get the other number

5 0
3 years ago
gary spent $350 on clothes. Jeans cost $50 each and shorts cost $15 each. if he bought 14 total items, how many jeans and shirts
mario62 [17]

Answer:

Step-by-step explanation:

6 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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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
What is 2x+5=y and 5x+4=y pls hurry
Sergeu [11.5K]

Answer:

( 1/3, 5 2/3) Or (0.33, 5.66)

Step-by-step explanation:

If you graph the 2 equations and see where they intersect, they will land on the answer.

4 0
3 years ago
How do I solve this​
Alina [70]

\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\displaystyle\sum \limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio} \end{cases} \\\\[-0.35em] ~\dotfill

\bf S_{20}=\displaystyle\sum \limits_{n=1}^{\stackrel{\stackrel{n}{\downarrow }}{20}}~\stackrel{\stackrel{a_1}{\downarrow }}{3}(\stackrel{\stackrel{r}{\downarrow }}{1.5})^{n-1}\implies S_{20}=3\left(\cfrac{1-1.5^{20}}{1-1.5} \right)\implies S_{20}=3\left(\cfrac{1-\stackrel{\approx}{3325.3}}{-0.5} \right) \\\\\\ S_{20}=3\left(\cfrac{-3324.3}{-0.5} \right)\implies S_{20}=3(6648.6)\implies S_{20}=19945.8

8 0
3 years ago
Read 2 more answers
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