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The sum of 3 consecutive odd numbers is 183 whats the second number im this sentence

Naddik [55]

Answer:

The second number is 61

Step-by-step explanation:

It's implied that the first integer is odd.

Since the consecutive intergers are odd, use an equation that looks like this:

(x+2) + (x+2) + (x+2) = 183 and simplify

*If you add 2 to an odd number, you get the next odd number. That's why I used 2*

3x + 6 = 183 → 3x = 177 → x = 59 which means the first integer is 59. The consecutive integers is 59, 61, 63.

6 0

2 years ago

I neep help please A.S.A.P

Oksi-84 [34.3K]

Hannah surveyed 1/9th of the population.
The sample wouldn't really be representative of the population, because of the
small sample size.
Based on the table, you could expect around 60% of the students to own a cat or a dog. This is because there are 30 students with a pet, and 20 without, which makes a percentage of 60% students that own a pet.

6 0

2 years ago

On a test, mean score was 70 and the standard deviation of the scores was 15. What percent of the test takers scored above 80

Anna007 [38]

Answer:

25%.

Step-by-step explanation:

When calculating percentile of a value within a sample population, you'll want to use what's called the z-score formula, notated as such:

z = (x – μ) / σ

Or, the z-score equals the difference of the x-value minus the mean (μ), divided by the standard deviation (σ). We can use the equation to sub in our numbers:

$z=\frac{80-70}{15}\\z=\frac{10}{15}\\\\z=\frac{2}{3} \\z=0.667$

Since we are looking for scores above 80, what we want to find is the area under the normal distribution curve that is to the right of our value of 80, or the pink area illustrated in the attached picture.

For this, we can use the standard Z table to look up our z-score and find what percent of our chart we've filled, to the left of our value. You can find Z tables in algebra textbooks or online, like this one: http://www.z-table.com/

In this instance, our z-score gives us a resulting Z table percentile between .7454 (z-score of 0.66) and .7486 (z-score of 0.67). This means that, in our normal distribution with a mean of 70 and a standard deviation of 15, the area to the left of the value of 80 is about 74.6%.

So, now we know that our value of 80 makes up about 74.6% of our curve from the left. To find the area to the right, we just subtract from the total area, which is 1 in a normally distributed curve. This gives us 25.4%. Since we're looking for the percent of test takes scoring above our value, I'll round that down to 25%.