Round 154,101 to the nearest ten.

Kazeer [188]

6x^2 - 2x + 1 is a quadratic formula from the form ax^2 + bx + c. This form of equation represents a parabola.
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.

In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.

I've added a picture of the parabola represented by this equation under the answer.

Hope this Helps! :)

4 0

3 years ago

Arrange these plants by height from shortest to tallest: 3.6m, .45m, 6.2m, 1.23m

jarptica [38.1K]

Answer:

0.45m , 1.23 m, 3.6m, 6.2m

Step-by-step explanation:

Remember to note the placement of the decimal points. The greater the number(s) leaning towards the left, the greater the number will be.

~<em>Senpai</em>

8 0

3 years ago

According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard devi

Mekhanik [1.2K]

Answer:

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is mean amount of inches of rain and $\sigma$ is the standard deviation.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Mean $\mu$ , standard deviation $\sigma$

n days:

This means that $s = \frac{\sigma}{\sqrt{n}}$

Applying the Central Limit Theorem to the z-score formula.

$Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?

The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is mean amount of inches of rain and $\sigma$ is the standard deviation.

5 0

2 years ago

Ms. irwin teaches four periods of weight training. she required each student in thr class to record the maximum amount of weight

son4ous [18]

Answer:

a

Step-by-step explanation:

Please mark brainliest

3 0

2 years ago

A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea

uysha [10]

Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n, the mean is $\mu*n$ and the standard deviation is $s = \sqrt{n}*\sigma$