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2 years ago

Round 3659 to the nearest thousand

Mathematics

2 answers:

Alexxandr [17]2 years ago

6 0

your answer is 4,000 ....... :))))))

Kryger [21]2 years ago

5 0

It is 4,000. :P
The 6 in the hundreds tells you to round up. 4 and under, keep it. 5 and up, raise it by one. :)

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Use Bayes' rule to find the indicated probability. The incidence of a certain disease on the island of Tukow is 4%. A new test h

lina2011 [118]

These are the events in the question above:

D - has disease

H - healthy (does not have disease)

P - tests positive 

It is the probability that a person has the disease AND tests positive divided by the probability that the person tests positive.

Sick, + [.04*.91] = .0364

Sick, - [.04*.09] = .0036 

Healthy, + [.96*.04] = 0.0384

Healthy, - [.96*.96] = .9216


.0364 / (.0364 + .0.0384) = 0.487

7 0

2 years ago

Lyn invested $7,000 into a investment paying 3% interest, compounded semi-annually, twice a year. After five years, how much wou

gayaneshka [121]

Answer:

Option B.) $8,123.79

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=5\ years\\ P=\$7,000\\ r=0.03\\n=2$

substitute in the formula above

$A=\$7,000(1+\frac{0.03}{2})^{2*5}$

$A=\$7,000(1.015)^{10}=\$8,123.79$

3 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

Solve this questions with steps

Alex_Xolod [135]

Provide better info thanks

4 0

2 years ago

Scores on a test have a mean of 70 and q3 is 83. the scores have a distribution that is approximately normal. find p90. (you wil

Hunter-Best [27]

Answer: Q3 represents 75%, meaning a z of ~0.67 80 - 70 is 10, so the standard deviations is ~14.9. 10 / 0.67 = 14.9 now find the z that represents a score of 90 90 - 70 is 20 20 / 14.9 = 1.34 from a z-table, a z of 1.34 represents a probability of ~90.99% meaning that there is about a 9.01% chance of getting a 90 or better.

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2 years ago