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wlad13 [49]
3 years ago
11

The amount of rainfall in january in a certain city is normally distributed with a mean of 4.3 inches and a standard deviation o

f 0.3 inches. find the value of the quartile q1. 1.1 4.2 4.1 4.5

Mathematics
1 answer:
Sholpan [36]3 years ago
7 0
We start by showing the position of the first quartile, Q1 on the normal distribution graph. Q1 happens at 25% of the data.

The graph is shown below

We are interested in the area to the left of Q1 and we know that the probability is 0.25

We can find out the value of z-score from the table as shown below 

The z-score when p=0.25 is 0.68

then changing this z-score into the actual value

-0.68= \frac{X-4.3}{0.3}
-0.204=X-4.3
X=4.096

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90 POINT QUESTION!!!
Aleks [24]

Answer: 9.19 ft

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the next trigonometric function.  

Sin α = opposite side / hypotenuse  

Where α is the angle of elevation of the ladder to the ground, the hypotenuse is the longest side of the triangle (in this case is the length of the ladder), and the opposite side (x) is the height of the top of the ladder above the ground.

Replacing with the values given:  

Sin 45 = x/ 13

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sin45 (13) =x  

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Feel free to ask for more if needed or if you did not understand something.  

7 0
3 years ago
What is the expression of (0.2)(14)
Natalija [7]
The expression is 2.8 
5 0
3 years ago
What is the number between 3/25 and the square root of 14
astra-53 [7]

Answer:

3.74165738677

Step-by-step explanation:

6 0
3 years ago
A/45 = 5/7 Solve for a
Charra [1.4K]

Answer:

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5 0
3 years ago
Read 2 more answers
If a cup of coffee has temperature 97°C in a room where the ambient air temperature is 25°C, then, according to Newton's Law o
FinnZ [79.3K]

Answer:

The  average temperature  is T_{a} = 81.95^oC

Step-by-step explanation:

From the question we are told that

    The temperature of the coffee after time t is   T(t) =  25 + 72 e^{[-\frac{t}{45} ]}

Now the average temperature during the first 22 minutes i.e fro 0 \to  22minutes is mathematically evaluated as

              T_{a} =  \frac{1}{22-0}  \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx

               T_{a} = \frac{1}{22} [25 t  +  72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.

             T_{a} = \frac{1}{22} [25 t  - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.

              T_{a} = \frac{1}{22} [25 (22)  - 3240e^{[-\frac{22}{45} ]}   - (- 3240e^{0} )]

            T_{a} = \frac{1}{22} [550  - 1987.12  +  3240]

          T_{a} = 81.95^oC

       

8 0
3 years ago
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