Math help!!! will be marking brainly

A cable company has a one time installation fee and then a monthly charge to use their service. The total cost of modeled by th

SpyIntel [72]

Answer:

130=monthly charge

119=one time installation fee

Step-by-step explanation:

119 is the y-intercept and in problems like these, it is always the one time fee. 130 is the monthly payment because it has an x, which could account for the number of months. For example , say six months passed, you could multiply 130 * 6 to get the monthly fee for six months. I hope that makes sense!

6 0

2 years ago

Please, please help. Answer the question, then explain how you got your answer. <br><br> a= -3

Mice21 [21]

3.5*(-3)///////3.5*(-3)//////all of this parts same

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