92 divided by 11 with remainder

He normal pulse rate of a 13-year-old is between 70 and 100 beats per minute. As part of a health program, the medical staff at

SVEN [57.7K]

From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.

Thus, interquatile range of the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10

Similarly, from the box plot, it can be seen that for grade 8 students,
The least value is 76 and the highest value is 97. The lower and the upper quartiles are 85 and 94 respectively while the median is 89.

Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9

The difference of the medians of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5

Therefore, the difference of the medians is about half of the interquartile range of either data set.

6 0

2 years ago

Read 2 more answers

the class kept track of rainy and sunny days. During the 54 days, classified rainy or sunny, the ratio of rainy days is 7 to 14.

Rom4ik [11]

There are 36 sunny days

6 0

2 years ago

Read 2 more answers

Mrs. Thomas has $71.00 to purchase bottles of juice for her class. If the bottles of juice cost $3.55 each, how many bottles can

shepuryov [24]

B) 20
$71.00÷$3.55=20

5 0

2 years ago

Marco and his friends bought game tokens for 15$ and three admission tickets to fun place whih expression could be used to repre

bearhunter [10]

T= cost of one ticket
f(t)=3t+15

a)3t+15

3 0

2 years ago

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n

devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

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

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

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.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so $\mu = 49.9, \sigma = 3.74$ .

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when $Z = 1.04, \mu = 49.9, \sigma = 3.74$ .

So

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

$1.04 = \frac{X - 49.9}{3.74}$

$X - 49.9 = 3.8896$

$X = 53.7896$

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

8 0

2 years ago