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Dahasolnce [82]
3 years ago
5

Calculate the angle of depression, x, if the boat is 750 ft. away from the light house and the light house is 300 ft high. Round

your answer to the nearest tenth of a degree.

Mathematics
1 answer:
I am Lyosha [343]3 years ago
8 0

Firstly , we need to draw diagram

we are given

opposite =300

adjacent =750

now, we can use tan formula

tan(x)=\frac{300}{750}

now, we can find x

x=tan^{-1}(\frac{300}{750})

we get

x=21.8  degree

so,

angle of depression is 21.8 degree.....Answer

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Find the equation of the exponential function represented by the table below: above^^^^
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Answer:

Step-by-step explanation:

We will use 2 coordinates from the table along with the standard form for an exponential function to write the equation that models that data. The standard form for an exponential function is

y=a(b)^x where x and y are coordinates from the table, a is the initial value, and b is the growth/decay rate. I will use the first 2 coordinates from the table: (0, 3) and (1, 1.5)

Solving first for a:

3=a(b)^0. Sine anything in the world raised to a power of 0 is 1, we can determine that

a = 3. Using that value along with the x and y from the second coordinate I chose, I can then solve for b:

1.5=3(b)^1. Since b to the first is just b:

1.5 = 3b so

b = .5

Filling in our model:

y=3(\frac{1}{2})^x

Since the value for b is greater than 0 but less than 1 (in other words a fraction smaller than 1), this table represents a decay function.

4 0
3 years ago
Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T
frosja888 [35]

Answer:

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

p_v =P(Z  

a) 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 FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

b) \chi^2 =\frac{10-1}{1.96} 4 =18.367  

p_v =P(\chi^2 >18.367)=0.0311

If we compare the p value and the significance level provided 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 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

\bar X=7.25 represent the sample mean  

s=1.2 represent the sample standard deviation

\sigma=1.4 represent the population standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test  

\alpha=0.05,0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 7.5  

Alternative hypothesis:\mu < 7.5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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".  

3) Calculate the statistic  

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

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

4)P-value  

Since is a left tailed test the p value would be:  

p_v =P(Z  

5) Conclusion  

Part a

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 FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

Part b

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=10 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =4 represent the sample variance obtained

\sigma^2_0 =1.96 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \leq 1.96

Alternative hypothesis: \sigma^2 >1.96

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{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 =\frac{10-1}{1.96} 4 =18.367

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2 >18.367)=0.0311

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

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided 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 1.96.

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3 years ago
What is 3(2x-6)-11=4(x-3)+6
navik [9.2K]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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