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3 years ago

11

Milanputs 1 4 ŽĨŚĞƌůĂǁŶͲŵŽǁŝŶŐŵŽŶĞLJŝŶƐĂǀŝŶŐƐĂŶĚƵƐĞƐ 1 2of the remaining money to pay back her sister. If she has $15 left, how

much did she have at first?

1 answer:

kirill115 [55]3 years ago

7 0

The same is 14 maybe you could try it

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1. In a class of 30 students, 5 made A's on the first test. If a student is selected at random to do a

ExtremeBDS [4]

Step-by-step explanation:

State you answer in the simplest form

6 0

3 years ago

in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru

melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

$\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298$

The 90% confidence interval for the proportion would be then

$\bf [0.77-z^*0.0298, 0.77+z^*0.0298]$

where $\bf z^*$ is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval $\bf [-z^*,z^*]$ is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

$\bf [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]$

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0

3 years ago

Can you help? I'm just too lazy to answer it lol

Usimov [2.4K]

Answer:

second answer should be corr3ect tell em if wrong

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

For an annual membership fee of $500, Mr.Bailey can join a country club that would allow him to play a round of golf for $35. Wi

mote1985 [20]

He would have to play 11 games. and 3 games plus a membership in order to equal the same amount of 605

6 0

3 years ago

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab

Sphinxa [80]

Answer: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

$\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5$

Using the poisson process, we get that

(a) less than or equal to 12

probability= $P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980$

(b) equal to 13

probability= $P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013$

(c) greater than 12

probability= $P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020$

(d) equal to 20

probability= $P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026$

(e) between 10 and 15, inclusively

probability= $P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318$

Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0

3 years ago