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Aliun [14]
3 years ago
6

It is estimated that 52% of drivers text while driving. How many people should a police officer expect to pull over until she fi

nds a driver NOT texting while driving?
Mathematics
1 answer:
sasho [114]3 years ago
4 0

Answer:

2

Step-by-step explanation:

Let

P = percentage of those that text and drive

S = percentage of those that do not text and drive

P + S = 1

S = 1 - P

S = 1 - 0.52

S = 0.48

The expected number would be:

1/0.48 = 2.08

Which is approximately 2.

Therefore the expected number of people the police officer would expect to pull over until she finds a driver not texting is 2.

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Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
In a bag of marbles 12% were blue 18% were green and the rest were purple if the bag has 150 marbles how many were blue or green
baherus [9]

Answer:

45

Step-by-step explanation:

12 % of the marbles are blue, and

18 % of the marbles are green, so

30 % of the marbles are either blue or green.

There are 150 marbles.

30 % × 150 = 0.30 × 150 = 45

The number of blue or green marbles is 45.

8 0
3 years ago
National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a deskt
Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, H_0 : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, <u><em>test statistics</em></u>  =  \frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }

                               =  2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0
2 years ago
How do I do this problem?
4vir4ik [10]
You have to put the number under 1 so you would get 1 over -27
3 0
2 years ago
Three out of every five students wore green on St. Patrick's Day. What percent of the students wore green?
aivan3 [116]
60% must be the right answer
6 0
3 years ago
Read 2 more answers
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