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Arada [10]
3 years ago
5

Find the mean for the given set of data. 18, 15, 22, 18, 25, 19, 23 18 19 20

Mathematics
2 answers:
Colt1911 [192]3 years ago
7 0

Answer:

20

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\left[\begin{array}{cc}Mean(Average):&20\\Median:&19\\Range:&10\\Mode:&18, \mathrm{\:appeared\:2\:times}\\\mathrm{\:Geometric\:Mean}:&19.740326987854\\Largest:&25\\Smallest:&15\\Sum:&140\\Count:&7\end{array}\right]

\mathrm{Sorted\:Data\:Set}: 15, 18, 18, 19, 22, 23, 25

enot [183]3 years ago
4 0

Answer:

20

Step-by-step explanation:

When you add them all up and divide by seven, you get 20

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Determine weather randomly picking a blue card from a bag containing all blue cards is impossible, unlikely, as likely as not, l
Anuta_ua [19.1K]

Answer:

The correct answer is certain with probability equal to 1.

Step-by-step explanation:

Probability is a mathematical framework which helps us to analyze chance of the outcome in a particular experiment. The value of probability is given by the ratio of the possible outcomes favorable to a certain experiment to the total outcomes.

We say an event is certain when the probability is 1 and the probability is zero when the event is uncertain.

Here the experiment is picking a blue card from a bag containing all blue cards.

Possible outcomes are all the cards colored blue in the bag.

Total outcomes are also all the blue cards in the bag.

∴ The value of probability is 1 as the event is certain because if we pick a card from the bag containing only blue cards, it would certainly give us a blue card.

5 0
2 years ago
To the nearest mile, what is the total distance that he ran?
ioda

Answer:

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Step-by-step explanation:

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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

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Ilia_Sergeevich [38]
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If h(z)=13z2–30z+27, use synthetic division to find h(3).
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Answer:

h=13z-30+ 27/z

Step-by-step explanation:

4 0
3 years ago
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