MATHEMATICS COLLEGE

Find the product: -5(-3)(-3)

Answers

Answer 1

Answer:

Answer:

-45

Step-by-step explanation:

Answer 2

Answer:

Answer:

((−5)(−3))(−3)

=−45

hope this was helpful!

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In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 241.4 sec. and a standard deviation of 57.59 sec. Use a 0.05 significance level and the accompanying minitab display to test the claim that the sample is from a population of songs with a mean great thatn 210 sec. What do these results suggest about the advice given in the manual.The mini tab displays the following:

One-sample T

Test of mu=210 vs.>210

N Mean St. Dev SE Mean 95% lower bound T p

40 241.40 57.59 9.11 226.06 3.45 0.001

A H0 u>210 sec. H1 u < 210sec

B H0 u=210 sec. H1 u < 210sec

C H0 u<210 sec. H1 u> 210sec

D H0 u=210 sec. H1 u> 210sec

Identify the test statistic:

T =

Identify the P-Value

P-value=

Stat the final conclusion that addresses the original claim. Choose from below:

A. Reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.
B. Fail to reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length great thatn 210 sec.
C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.
D. Fail to reject H0. There is sufficient evidence to support the claim tha tthe sample is from a population of songs with a mean lenght greater than 210 sec.

What do the results suggest about the advice given in the manual?

A. The results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.
B The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice
C. The results suggest that 241.4 seconds is the best song lenght.
D. The results are inconclusive because the average length of a hit song is constantly changing.

Answers

Answer:

D H0 u=210 sec. H1 u> 210sec

$t=(241.4-210)/((57.59)/(√(40)))=3.448$

$p_v =P(t_((39))>3.448)=0.000684$

C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.

Step-by-step explanation:

Data given and notation

$\bar X=241.4$ represent the sample mean

$s=57.59$ represent the sample standard deviation for the sample

$n=40$ sample size

$\mu_o =210$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is greater than 210 seconds, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210$

Alternative hypothesis: $\mu > 210$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=(241.4-210)/((57.59)/(√(40)))=3.448$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_((39))>3.448)=0.000684$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 210 seconds.

C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.

The Food and Drug Administration advises people against eating fish with mercury concentrations exceeding 1 part per million (ppm). Researchers wonder if the average mercury content of large-mouth bass in a certain state exceeds the limits and, thus, make such fish inedible. They randomly sample largemouth bass from randomly selected lakes in this state and record the amount of mercury in each fish. What is the correct parameter in this example

Answers

Answer:

The average amount of mercury in the population of large-mouth bass in this particular state

Step-by-step explanation:

Parameters are properties that describe entire populations rather than a sample. Therefore, since researchers are focused on the average mercury content of large-mouth bass, the correct parameter for the experiment would be the average amount of mercury in the population of large-mouth bass in this particular state.

Four research teams each used a different method to collect data on how fast a new iron skillet rusts. Assume that they all agree on the sample size and the sample mean (in days). Use the (confidence level; confidence interval) pairs below to select the team that has the smallest sample standard deviation.A. Confidence Level: 99.7%; Confidence Interval: 40 to 40
B. Confidence Level: 95%; Confidence Interval: 40 to 50
C. Confidence Level: 68%; Confidence Interval: 43 to 47
D. Confidence Level: 95%; Confidence Interval: 42 to 48

Answers

ConfidenceLevel: 95%; Confidence Interval: 42 to 48. Then the correct option is D.

How to interpret the confidence interval?

Suppose the confidence interval at P% for some parameter's values is given by x ± y.

That means that the parameter's estimated value is P% probable to lie in the interval

[x - y, x + y]

Four research teams each used a different method to collect data on how fast a new iron skillet rusts.

Assume that they all agree on the sample size and the sample mean (in days).

Use the (confidence level; confidence interval) pairs below to select the team that has the smallest sample standard deviation.

Then we have

48 - 42 = 6

Then we have

6/2=3

3 is smallest

Then the correct option is D.

Learn more about confidence intervals here:

brainly.com/question/16148560

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Answer:

D. Confidence Level: 95%; Confidence Interval: 42 to 48

Step-by-step explanation:

48-42=6

6/2=3

3 is smallest

I tested A and got it incorrect so D is the awnser

Which equation represents the line that passesthrough the point (1,5) and has a slope of -2?
1) y = -2x + 7
2) y = -2x + 11
3) y = 2x-9
4) y = 2x + 3

Answers

Answer:

Step-by-step explanation:

y - 5 = -2(x - 1)

y - 5 = -2x + 2

y = -2x + 7

Option 1 is the answer

A culture started with 5000 bacteria after three hours it grew 6500 bacteria predict how many bacteria will be present after 13 hours round your answer to the nearest whole number

Answers

Answer:

15595 bacteria will be present after 13 hours.

Step-by-step explanation:

Continuous population growth:

The continuous population growth model, for the population after t hours, is given by:

$P(t) = P(0)e^(rt)$

In which P(0) is the initial population and r is the growth rate.

Started with 5000 bacteria

This means that $P(0) = 5000$

$P(t) = 5000e^(rt)$

After three hours it grew 6500 bacteria:

This means that $P(3) = 6500$ . We use this to find r.

$P(t) = 5000e^(rt)$

$6500 = 5000e^(3r)$

$e^(3r) = (65)/(50)$

$\ln{e^(3r)} = \ln{(65)/(50)}$

$3r = \ln{(65)/(50)}$

$r = \frac{\ln{(65)/(50)}}{3}$

$r = 0.0875$

$P(t) = 5000e^(0.0875t)$

How many bacteria will be present after 13 hours?

This is P(13). So

$P(13) = 5000e^(0.0875*13) = 15594.8$

Rounding to the nearest whole number

15595 bacteria will be present after 13 hours.

John made a lump sum deposit of $ 6,300 in an account that pays 6.5% per year. Findthe value (maturity value) of his account after 5 years. How much is the interest?

Answers

The maturity value in John's account is $8347.5.

Given that, principal =$6300, rate of interest =6.5% and the time period =5 years.

What is the maturity value?

Maturity value is the amount due and payable to the holder of a financial obligation as of the maturity date of the obligation. The term usually refers to the remaining principal balance on a loan or bond. In the case of a security, maturity value is the same as par value.

Now, S.I. = (P×T×R)/100

= (6300×5×6.5)/100

= 2047.5

So, interest =$2047.5

Maturity value = Principal + Interest

= 6300+2047.5

= $8347.5

Therefore, the maturity value in John's account is $8347.5.

Learn more about the maturity value here:

brainly.com/question/2496341.

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Answer: $8,347.50 is the value of his account after five years.