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vredina [299]
1 year ago
6

The first time a census was taken in Esinville, the population was 12,621. Each year after that, the population was about 6% hig

her than it was the previous year. Sketch the graph that represents the situation. Let x represent the number of years since the first census was taken. What are the coordinates of the point that contains the ­y-intercept?
Although answers are nice, I need step by step to help me understand how to solve this. Its exponential growth I believe?

Mathematics
1 answer:
Sedbober [7]1 year ago
8 0
Step-by-step
Population was 12.621 when it was measured

Next year it rose by 6%, so add that 6% to 100% and multiply with population measured

12.621 × (100%+6%) = 12.621 × 106% = 13.379,24

Since a man cant be 0,24, we count that as +1

13.379,24 = 13.379 + 1 = 13.380

That was for a year after measurement, now let's calculate for another year
Now, add those 6% to a new population number

13.380 × (100%+6%) = 13.380 × 106% = 14.182,8

since 0,8 isn't a man, instead we add again +1

14.182,8 = 14.182 + 1 = 14.183

now let's do it for another year after last, by adding those 6% to a newest population number

14.183 × (100%+6%) = 14.183 × 106% = 15.033,98

round it again to +1, since 0.98 isn't a man

15.033,98 = 15.033 + 1 = 15.034

that is it, now graph is in the picture.

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The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi
SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so n = 1210.

The mean of the sample is 1030.

The probability of a success is \pi = \frac{1030}{1210} = 0.8512.

The standard deviation of the sample is s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when X = 1030.

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

\mu = 0.82(1210) = 992.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{1030 - 992.2}{12.38}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989.This means that the county result is unusually high.

4 0
2 years ago
Every possible cross section of a three dimensional figure is a circle what is that figure
Korolek [52]
A sphere. It is a perfectly circular and even shape, so all cross sections would be circles. 
5 0
2 years ago
Read 2 more answers
Evaluate 4-0.25g+0.5h4−0.25g+0.5h4, minus, 0, point, 25, g, plus, 0, point, 5, h when g=10g=10g, equals, 10 and h=5h=5h, equals,
Readme [11.4K]

I believe the correct given equation is in the form of:

4 – 0.25 g + 0.5 h

Now we are to evaluate the equation with the given values:

g = 10 and h = 5

What this actually means is that to evaluate simply means to calculate for the value of the equation by plugging in the values of the variables. Therefore:

4 – 0.25 g + 0.5 h = 4 – 0.25 (10) + 0.5 (5)

4 – 0.25 g + 0.5 h = 4 – 2.5 + 2.5

4 – 0.25 g + 0.5 h = 4

 

Therefore the value of the equation is:

4

6 0
1 year ago
Read 2 more answers
A bag holds 11 beads. 7 are red, the rest are white. Two beads are taken at random from the bag. What is the probability that on
Flura [38]

Answer:

28/55

Step-by-step explanation:

The probability of taking a red bead first and a white bead second is:

P(red, white) = (7/11) (4/10)

P(red, white) = 14/55

The probability of taking a white bead first and a red bead second is:

P(white, red) = (4/11) (7/10)

P(white, red) = 14/55

So the probability that one bead of each color is taken is:

P = P(red, white) or P(white, red)

P = 14/55 + 14/55

P = 28/55

4 0
1 year ago
Read 2 more answers
An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "
FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

\large X_{sentences} the mean of the “sentences” group

\large S_{sentences} the standard deviation of the “sentences” group

\large X_{intentional} the mean of the “intentional” group

\large S_{intentional} the standard deviation of the “intentional” group

Then, we can calculate by using the computer

\large X_{sentences}=28.75  

\large S_{sentences}=3.53553

\large X_{intentional}=31.625

\large S_{intentional}=1.40788

\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}

So, the standard error of the difference is

\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of  95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the \large t_{0.05} which is a value of t such that the area under the Student's t distribution  outside the interval \large [-t_{0.05}, +t_{0.05}] is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

\large t_{0.05}=2.365

Now we can compute our confidence interval

\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0
2 years ago
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