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Lapatulllka [165]
3 years ago
14

I need help on this.

Mathematics
2 answers:
den301095 [7]3 years ago
8 0

Answer:

yes

Step-by-step explanation:


dem82 [27]3 years ago
3 0

Answer:


Step-by-step explanation:

Yes


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The distance a car has traveled can be represented by the equation d = 1.4m, where d represents the distance in miles and m repr
ValentinkaMS [17]

Answer:

It is the velocity of the car, the speed.

5 0
3 years ago
Candice is preparing for her final exam in Statistics. She knows she needs an 74 out of 100 to earn an A overall in the course.
Salsk061 [2.6K]

Answer:

Hence the correct option is option b) Yes, the upper level of one standard deviation is 72.

A score of 74 is not within one standard deviation of the mean.

Step-by-step explanation:

Here the given details are,

Mean = 68  

SD = 4  

Distribution is normal.  

Z-score for x = 74 is given as below:  

Z = (X - mean)/SD\\Z = (74 - 68)/4\\Z = 1.5  

So, the score of 74 is 1.5 standard deviations from the mean.  

Mean + 1\timesSD = 68 + 1\times4 = 72Mean - 1\timesSD = 68 - 1\times4 = 64  

Therefore the score is not lies between 64 and 72.  

Yes, the upper level of one standard deviation is 72.  

3 0
3 years ago
Yesterday,
Mashutka [201]

Answer:

45

Step-by-step explanation:

5 divided by 8, multiply by 72

=(5÷8)72

=(0.625)72

=45

6 0
3 years ago
g A population is infected with a certain infectious disease. It is known that 95% of the population has not contracted the dise
trasher [3.6K]

Answer:

There is approximately 17% chance of a person not having a disease if he or she has tested positive.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = a person has contracted the disease.

+ = a person tests positive

- = a person tests negative

The information provided is:

P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01

Compute the missing probabilities as follows:

P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99

The Bayes' theorem states that the conditional probability of an event, say <em>A</em> provided that another event <em>B</em> has already occurred is:

P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}

Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:

P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}

              =\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667

So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.

As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.

7 0
3 years ago
What is the solution to the equation 4 + = 6?
Anna007 [38]

Answer:

2

4 + 2 = 6

Hope this helps!! Pls mark me brainliest!!

4 0
3 years ago
Read 2 more answers
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