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3 years ago

Realiza tres ejemplos de probabilidad subjetiva

Mathematics

1 answer:

Lemur [1.5K]3 years ago

5 0

ojala te sirva

es niño o niña?: La abuela de un bebe que esta por nacer asegura que el futuro bebé será una niña y no un niño, la abuela se basa en los antojos de la madre, la forma del estómago de la madre y otras características que ella ha podido observar durante el embarazo.

te comparto imagenes ok

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A high school athlete ran the 100 meter sprint in 13.245 seconds. what is the time rounded to the nearest tenth. Award is brainl

monitta

If you round 13.245 seconds to the nearest tenth, the answer is 13.2 seconds.
Hope this helped!

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3 years ago

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The lines shown below are perpendicular. If the green line has a slope of

klasskru [66]

Answer: 4

Step-by-step explanation:

Perpendicular lines have reciprocal slopes

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3 years ago

Darren wins a coupon for $4 off the lunch special for each of five days he pays $75 for his 5 lunch specials write and solve an

Valentin [98]

The original price for one lunch special is $19.

Explanation

The original price for one lunch special is 'p' dollar.

He wins a coupon for $4 off for each of five days. That means , he needs to pay $(p-4)$ dollar each day.

So, the total amount needed to pay for 5 days $= 5(p-4)$ dollar

Given that, he pays $75 for his 5 lunch specials. So the equation will be.....

$5(p-4)= 75\\ \\ 5p-20=75\\ \\ 5p= 75+20\\ \\ 5p= 95\\ \\ p=\frac{95}{5}=19$

So, the original price for one lunch special is $19.

8 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

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3 years ago

What is the average rate of change? f(x)=x^2+3 From x=0 to x=3

SpyIntel [72]

Answer:

Step-by-step explanation:

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3 years ago