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3 years ago

Need some help someone please?

Mathematics

2 answers:

zlopas [31]3 years ago

8 0

I am pretty sure that the answer is 15+ n or it could be 15n

Svet_ta [14]3 years ago

7 0

The answer is c 15 plus m

You might be interested in

How can you determine if you need to use a combination or permutation to count the number of outcomes? Which will usually have m

Nadya [2.5K]

Answer with Step-by-step explanation:

Permutation : It is an arrangement of r elements out of n elements.

Combination : it is a selection of r element out of n elements .

Suppose we have a set

S={1,2,3}

If two elements are taken at a time then

Using permutation formula

Total number of outcomes= $3P_2$

Total number of outcomes= $\frac{3!}{(3-2)!}$

Total number of outcomes=3!= $3\times 2\times1=6$

Using combination formula

$\binom{n}{r}=\frac{n!}{r!(n-r)!}$

Total number of outcomes= $\binom{3}{2}=\frac{3!}{2!1!}$

Total number of outcomes= $\frac{3\times2!}{2!}$

Hence, total number of outcomes=3

Total number of outcomes determined by permutation have more outcomes.

Because permutation is an arrangement of elements therefore, it consider order of arrangement of element but combination is a selection of elements it does no consider order of elements

Arrangements of two elements out of 3 elements

{1,2},{2,3},{2,1},{3,2},{1,3},{3,1}

By using combination if two elements taken at a time then combination

{1,2},{2,3},{1,3}

4 0

3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]