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Ivahew [28]
3 years ago
11

Suppose a sample of 1390 suspected criminals is drawn. Of these people, 514 were captured. Using the data, estimate the proporti

on of people who were caught after being on the 10 Most Wanted list. Enter your answer as a fraction or a decimal number rounded to three decimal places.
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
4 0

Answer:

The estimate for the proportion of people who were caught after being on the 10 Most Wanted list is of 0.37.

Step-by-step explanation:

Estimate of the proportion of people who were caught after being on the 10 Most Wanted list.

Number of people captured divided by the size of the sample.

We have that:

514 of 1390 people were captured. So

514/1390 = 0.37

The estimate for the proportion of people who were caught after being on the 10 Most Wanted list is of 0.37.

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A set of chemistry exam scores are normally distributed with a mean of 70 points and a standard deviation
Fed [463]

Answer:

38.117%

Step-by-step explanation:

Firstly, what you do is to calculate the z-score of both points

Mathematically;

z-score = (x-mean)/SD

where mean = 70 and SD = 5

For 68, z-score = (68-70)/5 = -2/5 = -0.4

for 73, z-score = (73-70)/5 = 3/5 = 0.6

The probability we are to calculate ranges from

P(-0.4<z<0.6) = P(z<0.6) - P(z<-0.4)

it is at this point that we make use of z score table

= 0.38117 which is 38.117%

4 0
3 years ago
If the top of a 4.50 m ladder reaches twice as far up a vertical wall as the foot of the
Free_Kalibri [48]

Answer:

4.02 meters.

Step-by-step explanation:

In the diagram, the length of the ladder is |AC|.

If the foot of the ladder is x meters from the base of the ladder

Then the distance of the ladder up the wall, AB=2x meters.

Using Pythagoras Theorem

|AC|^2=|AB|^2+|BC|^2\\4.5^2=(2x)^2+x^2\\4.5^2=4x^2+x^2\\20.25=5x^2\\$Divide both sides by 5\\x^2=20.25\div 5\\x^2=4.05\\x=\sqrt{4.05}=2.01 feet

Therefore, the distance of the ladder up the wall,

AB=2 X 2.01 =4.02 meters.

5 0
3 years ago
Read 2 more answers
determine any data values that are missing from the table assuming that the data represent a linear function
Pavel [41]

Answer:

The answer is B

Step-by-step explanation:

This is because if you take a look at the function table you are imputing 1 each time. So if you start with 1 in X and add 1 more to it you get 2 in y.

7 0
4 years ago
An article in the Journal of Materials Engineering (Vol 11, No. 4, 1989, pp. 275-282) reported the results of an experiment to d
RSB [31]

Answer:

Step-by-step explanation:

Hello!

You have two variables of interest

X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.

X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.

a)

To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.

If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test

If the population variances are different, σ₁²≠σ₂², you have to use Welch's t-test

Using α: 0.05

The normality test for X₁ shows a p-value of 0.7449 ⇒ You can assume it has a normal distribution.

The normality test for X₂ shows a p-value of 0.9980 ⇒ You can assume it has a normal distribution.

The F-test for variance homogeneity shows a p-value of 0.6968 (H₀:σ₁²=σ₂²) ⇒You can assume both population variances are equal.

b) and c)

You need to test if both population means are the same, the hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{8*4.28+5*5.62}{9+6-2} }= \sqrt{4.7953}= 2.189= 2.19

t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } = \frac{(16.36-11.48)-0}{2.19*\sqrt{\frac{1}{9} +\frac{1}{6} } } = 4.23

The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:

P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]=  0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001

The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.

I hope it helps!

6 0
3 years ago
Geometry question need help
Schach [20]
For my calculations the answer is 5
6 0
3 years ago
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