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raketka [301]
3 years ago
8

How do you find slope???!!! 0.0

Mathematics
1 answer:
dexar [7]3 years ago
3 0
My teacher gave me the same bellwork :3
The coordinate plane is 1 over 1 and the box is 4 over 1
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What is spark addition?
Arturiano [62]

Answer:

spark addition is a left to right addition

3 0
2 years ago
Marine scientists categorize signature whistles of bottlenose dolphins by typelong dash—type ​a, type​ b, type​ c, etc. In one s
r-ruslan [8.4K]

Answer:

The 95​% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 185, \pi = \frac{100}{185} = 0.5405

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 - 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.4687

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 + 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.6123

The 95​% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

3 0
3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
3 years ago
I NEED HELP ON THIS ONE PLEASEEEEE HELP MEE!!!!!!!!!!!
mojhsa [17]
A+b+c=28x+40=180
—>28x=140
—>x=140/28=5
c=9x-3
—>c=42
7 0
3 years ago
I need help please…thank youuu!!!
DochEvi [55]

Answer:

I will help ukshagaakavlabsbakakahsjskmdlslsbsbs skshsvs

4 0
3 years ago
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