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Marat540 [252]
3 years ago
6

The expression P(z > –1.53) represents the area under the standard normal curve above a given value of z. Find P(z > –1.53

).
Mathematics
1 answer:
Damm [24]3 years ago
6 0
We can only look up the value in the normal distribution table if the probability of z is LESS than a certain value AND the value is positive.
If the probability of z is MORE than the value, then you need to do 1 minus the probability.
I.e. 1 - 0.93699
[0.93699 is found in the normal distribution table, searching for 1.53]
However, since the value (-1.53) is also negative, you need to do 1 minus AGAIN. 
I.e. 1 - 0.93699

Since you have to do both, the answer should be obtained as follows:
1 - (1 - 0.93699) = 0.93699

Sorry for the terrible explanation :/ 
And apologies if anything is incorrect. Hope this helped though!
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A vertical pole 5ft long casts a shadow of 2ft at the same time a nearby tree casts a shadow of 10ft, how y’all is the tree?
SSSSS [86.1K]

Answer: The tree is 4 feet tall.

Step-by-step explanation: 5/2=x/10

2 x 10=20

20/5=4

5 0
3 years ago
The table shows the different types of shoes available at a family shoe store.
Wittaler [7]

Answer:

about 48.6%

Step-by-step explanation:

20 women's shoes are athletic.

35 women's shoes are formal.

52 women's shoes are casual.

The total number of women's shoes is 107.  So the probability that a randomly selected one is casual is:

P = 52/107

P ≈ 48.6%

6 0
3 years ago
Line r cuts a pair of parallel lines. One of the eight angles created measures 90°. Which statements about the angles are true?
MrRa [10]

Answer:

Step-by-step explanation:

if two parallel lines are cut by a traversal, the alternate exterior angles are congruent. If two lines are cut by a traversal and the alternate exterior angles are congruent, the lines are parallel. Corresponding Angles: The name does not clearly describe the "location" of these angles

4 0
4 years ago
The data show the traveler spend- ing in billions of dollars for a recent year for a sample of the states. Find the range, varia
Svetllana [295]

Solution :

Given data :

20.1     33.5     21.7      58.4     23.2     110.8     30.9

24.0    74.8     60.0

n = 10

Range : Arranging from lowest to highest.

20.1,   21.7,   23.2,    24.0,   30.9,    33.5,    58.4,    60.0,    74.8,   110.8

Range = low highest value - lowest value

           = 110.8 - 20.1

           = 90.7

Mean = $\frac{\sum x}{n}$

         $=\frac{20.1+21.7+23.2+24.0+30.9+33.5+58.4+60.0+74.8+110.8}{10}$

          $=\frac{457.4}{10}$

         $=45.74$

Sample standard deviation :

$S=\sqrt{\frac{1}{n-1}\sum(x-\mu)^2}$

$S=\sqrt{\frac{1}{10-1}(20.1-45.74)^2+(21.7-45.74)^2+(23.2-45.74)^2+(24.0-45.74)^2+(30.9-45)^2}$  

      \sqrt{(33.5-45.74)^2+(58.4-45.74)^2+(60.0-45.74)^2+(74.8-45.74)^2+(110.8-45.74)^2}

$S=\sqrt{\frac{1}{9}(657.4+577.9+508.0+472.6+220.2+149.8+160.2+203.3+844.4+4232.8)}$$S=\sqrt{\frac{1}{9}(8026.96)}$

$S=\sqrt{891.88}$

S = 29.8644

Variance = S^2

               =(29.8644)^2

               = 891.8823

6 0
3 years ago
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