Express the quantity of water flowing into a tank as a linear function of time. Then identify the quantity of water in the tank

Answer:

Refer to the detailed explanation below

Step-by-step explanation:

* Let’s organize the procedure to bisect angle ABC

- Angle ABC has vertex at point B and rays BA and BC

# Step 1:

- Center the compass at point B and draw an arc that crosses rays BA and BC ⇒ fifth step

# Step 2:

- Mark point D where the arc intersects ray BA and point E where it meets ray BC ⇒ third step

# Step 3:

- With the compass on point D, create an arc inside the angle ⇒ sixth step

# Step 4:

- Using the same compass width, set it on point E and draw an arc inside the angle ⇒ second step

# Step 5:

- Let point F be the intersection of the arcs from points D and E ⇒ fourth step

# Step 6:

- Draw a ray from point B through point F ⇒ first step

* This ray divides angle ABC into two equal parts

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176

Answer:

1. 3.767

2. 0.145

Step-by-step explanation:

Define X as the exam scores and Y as the number of drinks.

X     Y   X-Xbar    Y-Ybar   (X-Xbar)(Y-Ybar)    (X-Xbar)²       (Y-Ybar)²    

75    5    -2.3          2.3          -5.29                      5.29              5.29

92    3     14.7         0.3           4.41                       216.09           0.09

84    2     6.7         -0.7           -4.69                     44.89             0.49

64    4     -13.3        1.3           -17.29                     176.89           1.69

64    2     -13.3       -0.7           9.31                       176.89           0.49

86    7     8.7           4.3           37.41                     75.69            18.49

81     3     3.7           0.3           1.11                         13.69             0.09

61     0    -16.3        -2.7           44.01                     265.69          7.29

73    1      -4.3         -1.7            7.31                        18.49             2.89

93    0    15.7         -2.7           -42.39                    246.49          7.29

sumx=773, sumy=27, sum(x-xbar)(y-ybar)= 33.9, sum(X-Xbar)²= 1240.1,sum(Y-Ybar)²= 44.1

Xbar=sumx/n=773/10=77.3

Ybar=sumy/n=27/10=2.7

1.

Cov(x,y)=33.9/9

Cov(x,y)=3.76667

Thus, the sample covariance of exam scores and energy drink consumption is 3.767

2.

Cor(x,y)=r=33.9/233.85553

Cor(x,y)=r=0.14496

The sample correlation coefficient for the relationship between exam scores and energy drink consumption is 0.145.

Parce qu'en connaissant les techniques pour identifier les nombres rationnels, vous saurez automatiquement quand un nombre n'est pas rationnel.