I need helpppppppppppppppppppppppppppppp

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

6 0

Answer:

25 rooms

Step-by-step explanation:

Hotel rooms = 3 + 5 + 8 + 6 + 2 + 1 = 25

You might be interested in

consider the graph of the following quadratic equation. the equation of the quadratic function represented by the graph is y = a

dmitriy555 [2]

For x values 1 unit either side of the vertex, the value of y is 2 more than it is at the vertex.
a = 2

_____
More formally, The graph appears to go through the point (4, 1). Substituting these values for (x, y) in the equation, we get
1 = a(4-3)² -1 . . . . substitute for x and y
2 = a . . . . . . . . . . add 1 and simplify

7 0

3 years ago

Estimate the square root to the nearest integer. -51

marissa [1.9K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Data from the Centers for Disease Control and Prevention indicate that weights of American adults in 2005 had a mean of 167 poun

sesenic [268]

Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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