MATHEMATICS COLLEGE

In January 2005, the population of California was 36.8 million and growing at an annual rate of 1.3%. Assume that growth continues at the same rate. a) By how much will the population increase between 2005 and 2030

Answers

Answer 1

Answer:

Answer:

11,960,000 populations.

Step-by-step explanation:

Population of California at year 2005 = 36.8 million

If the population is growing at annual rate of 1.3%, then yearly increment will be:

1.3% of 36.8 million

= 1.3% of 36,800,000

= 1.3/100 * 36,800,000

= 1.3 * 368,000

= 478,400

The number of yeas between 2005 and 2030 is 25years

The population increase between 2005 and 2030 will be 25 * 478,400

= 11,960,000

Hence the population would have increased by 11,960,000 populations between 2005 and 2030

Answer 2

Answer:

Final answer:

Using the compound interest formula commonly used in mathematics, the population of California is expected to increase by approximately 13.04 million between 2005 and 2030, assuming an annual growth rate of 1.3%.

Explanation:

The problem here is related to compound interest in mathematics. Here, the population of California is growing annually at a rate of 1.3%, which means it's compounding, much like interest in a bank. To calculate the growth in population from 2005 to 2030, or 25 years, we can use the formula for compound interest which is P(1 + r/n)^(nt), where P is the initial population, r is the annual growth rate, n is the number of times the population grows per year, and t is the time in years.

In this case, P is 36.8 million, r is 1.3% or 0.013, n is 1 (since the population grows once a year), and t is 25 (the number of years from 2005 to 2030). If we plug these values into the formula, we get: 36.8(1 + 0.013/1)^(1*25).

This simplifies to a population of approximately 49.84 million in 2030. Therefore, the population increase over those 25 years is: 49.84 million - 36.8 million = 13.04 million.

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Which solution matches both equations ?

Answers

I believe the answer is C I’m really sorry if it’s wrong.

The filling variance for boxes of cereal is designed to be .02 or less. A sample of 41 boxes of cereal shows a sample standard deviation of .16 ounces. Use α = .05 to determine whether the variance in the cereal box fillings is exceeding the design specification.

Answers

Answer:

$\chi^2 =(41-1)/(0.02) 0.0256 =51.2$

$p_v =P(\chi^2_(40) >51.2)=0.1104$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(51.2,40,TRUE)"

If we compare the p value and the significance level provided 0.05 we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 0.02, so there is no a violation of the specifications.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=41$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.16^2 =0.0256$ represent the sample variance obtained

$\sigma^2_0 =0.02$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is higher than the standard of 0.02, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 0.02$

Alternative hypothesis: $\sigma^2 >0.02$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =(n-1)/(\sigma^2_0) s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =(41-1)/(0.02) 0.0256 =51.2$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case $df = 41-1=40$ . And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2_(40) >51.2)=0.1104$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(51.2,40,TRUE)"

Conclusion

Whoever answers correctly gets brainlist

Answers

Answer:

So now D would be at (-4 , 2)

Step-by-step explanation:

From D, you can calculate every other letter’s location. Yeah.

Answer:

Please mark me as brainliest!

Step-by-step explanation:

The coordinates:

D: ( 2,6 )

A: ( 2, 1 )

B: ( 5, 1 )

C: ( 5, 6 )

You want to translate this down 4 units and left 6 units.

So what you would do is this:

Formula = ( x - 6 , y - 4 )

D: ( -4, 2 )

A: ( -4, -3 )

B: ( -1, -3 )

C: ( -1, 2 )

I need help with math

Answers

How can I help you?

A deck of cards contains RED cards numbered 1,2,3, BLUE cards numbered 1,2,3,4,5,6, and GREEN cards numbered 1,2. If a single card is picked at random, what is the probability that the card is GREEN AND has an ODD number? Provide the final answer as a fraction.

Answers

Answer:

8/15 because its 8 odd numbers out of 15 numbers all together

Which graphs represent functions?

Answers

i would say the answer is c. only graph d ,, think of the line test to see if a graph is a function or not

Final answer:

Graphs that represent functions have one input corresponding to one output. Examples include straight lines, parabolas, and sine waves.

Explanation:

Graphs that represent functions are those in which every input has exactly one output. In other words, there can only be one value of y for each value of x. For example, a straight line, a parabola, or a sine wave are graphs that represent functions.

On the other hand, graphs that do not represent functions may have one input value mapping to multiple output values or no output values at all. Examples of such graphs include circles, ellipses, or a graph with one vertical line intersecting it at multiple points.

It's important to note that in a function, the vertical line test can be used to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, then the graph does not represent a function.

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