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mihalych1998 [28]
3 years ago
8

The box plot is based on the? mean median mode range

Mathematics
1 answer:
Westkost [7]3 years ago
3 0

PLEASE GIVE BRAINLIST

By reading this ithink the answer is B=median

A boxplot is a standardized way of displaying the dataset based on a five-number summary: the minimum, the maximum, the sample median, and the first and third quartiles. Minimum : the lowest data point excluding any outliers.

hope this helped

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If the students finished their assignment at 1:30 and they had been working for an hour and 45 minutes, what time did they start
bekas [8.4K]

Answer:

11:15

Step-by-step explanation:

7 0
3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
On the first day of a road trip, Roberto traveled 213 miles from home. On the second day, he drove 149 miles farther. How many m
jek_recluse [69]

The answer would be 362 miles

add the number of miles from day one and day two

4 0
2 years ago
Kyle and Mark started at the same location. Kyle traveled 5 miles east, while Mark traveled 3 miles west. How far apart are they
marishachu [46]

Answer:

8 miles apart lol

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Today Ian wants to run less than 7/12 mile which of the following distances is less than 7/12?
dybincka [34]

A fraction with a denominator of 4 that is less than 7/12 miles is 1/4.Let the fraction be  x/4. Since this fraction must be less than 7/12 miles, we can write the inequality,  x/4 < 7/12.We now solve the inequality for x to obtain,  x<7/12(4).

This simplifies to x<7/3

This implies that x<2 1/3

This is distance so x is positive.

So we can choose x=1 or x=2.

Therefore the possible fractions include 1/4 or 2/4.

In the simplest form, the most appropriate fraction is 1/4





6 0
3 years ago
Read 2 more answers
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