Decimal rounded to the nearest hundredth is 6.32. This decimal is greater than 6.32. What could this decimal be?

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

kodGreya [7K]

y^2 (y + 2) + 16(y + 2)

(y+2) (y^2+16)

3 0

3 years ago

Marvin used a customary unit of measure to estimate the weight of a watermelon he grew in his backyard what could the estimated

Andrej [43]

Answer:

Usually pounds (lb) .About an an average of 20 to 25 pounds (lbs)

Step-by-step explanation:

Since Marvin uses Customary units, and considering the Watermelon.

Pounds, He could estimate based on average of 20 to 25 pounds.

He could choose another customary units, but it wouldn't be practical and unusual to say a watermelon of 160 quarters, also it would leave all the listeners without parameters.

So, definitely pounds.

3 0

3 years ago

Concerns about the climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g. from r

natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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 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 $s = \frac{\sigma}{\sqrt{n}}$ .