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Stells [14]
2 years ago
8

If the probability is 0.54 that Stock A will increase in value during the next month and the probability is 0.68 that Stock B wi

ll increase in value during the next month, what is the greatest possible value for the probability that neither of these two events will occur.
Mathematics
1 answer:
Aloiza [94]2 years ago
3 0

P(A) =0.54

P(B)= 0.68

P'(A)= 1-0.54 = 0.46

P'(B)= 1- 0.68 = 0.32

The probability of neither of both event will occur:

= P'(A)×P'(B)

=0.46 × 0.32

=0.1472

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Step-by-step explanation:

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Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in
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Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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2 years ago
A rectangular prism has a length of 412 millimeters, a width of 412 millimeters, and a height of 6 millimeters. Sally has a stor
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2 years ago
The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of vi
baherus [9]

Answer: 2185

Step-by-step explanation:

Let p be the proportion of visitors are campers.

Given : The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state.

The prior proportion of visitors are campers : p=0.35

Allowable error : E= 2%= 0.02

We know that the z-value for 95% confidence = z_c=1.96

Then by Central Limit Theorem , the required sample size would be :

n=p(1-p)(\dfrac{z_{c}}{E})^2

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Simply , we get

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Answer:

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Step-by-step explanation:

Okay first you multiply length by width by height to get the volume which is:

15*8*6= 720

Braking it down:

15*8= 120

120*6= 720

So then you divide 720 by 3 because the 720 is the total value of the box but we just want to know what the volume of how much dirt is there:

720/3= 240 (Check on a calculator if you want to ;D)

So your answer is 240 cubic inches of soil in the flower box

<em>Hope this helps :D </em>

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