Help ASAP

A car used 1 1/5 gallons to travel 63 miles. At what rate does the car use gas ,in miles per gallon

love history [14]

Answer:

52.2 miles per gallon

Step-by-step explanation:

63/1 / 6/5=52.2

7 0

3 years ago

Read 2 more answers

PLEASE HELP ME ILL GIVE BRAINLIEST

Fed [463]

55 -10 = 452x2 = 47x7 = 494x4 = 1621 + 1 = 2277- 10 = 67x+30=100 = 70 You gave the answers. ???

5 0

3 years ago

The mean of a sequence of n numbers is m. If we split the sequence into two sequences of lengths n1 and n2 and compute their mea

Furkat [3]

The mean of a sequence of numbers is the average.

The true statement is: $\mathbf{mn = m_1 \times n_1 + m_2 \times n_2}$

The given parameters are:

$\mathbf{Mean = m}$

$\mathbf{Size = n}$

The mean of a dataset is calculated as:

$\mathbf{Mean = \frac{\sum x}{Size}}$

So, we have:

$\mathbf{m = \frac{\sum x}{n}}$

Multiply both sides by m

$\mathbf{\sum x = mn}$

When the sequence is split into two, we have:

$\mathbf{\sum x_1 = m_1\times n_1}$

$\mathbf{\sum x_2 = m_2\times n_2}$

Where:

$\mathbf{\sum x_ = \sum x_1 + \sum x_2}$

So, we have:

$\mathbf{mn = m_1 \times n_1 + m_2 \times n_2}$

Hence, the true statement is: $\mathbf{mn = m_1 \times n_1 + m_2 \times n_2}$

Read more about mean and averages at:

brainly.com/question/16217700

8 0

3 years ago

Based on data from a college, scores on a certain test are normally distributed with a mean of 1530 and a standard deviation of

harkovskaia [24]

Answer:

0.73% of the scores are greater than 2317.

14.46% of the scores are less than 1190.

38.23% of the scores are between 1351 and 1673.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 1530, \sigma = 322$