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vfiekz [6]
3 years ago
7

Angle ABD is 90 what is the measure of ABC

Mathematics
1 answer:
Anarel [89]3 years ago
3 0

Answer:

90 I am guessing. if it's like a square or rectangle

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What is the degree of the polynomial? *<br> 2x+y+4x5y + 2xºy<br> W
Airida [17]

Answer:

Oh this is easy it is 6 so yea I'm right

7 0
3 years ago
Four lawn sprinkler heads are fed by a 1.9-cm-diameter pipe. The water comes out of the heads at an angle of 35° above the horiz
klemol [59]

Answer:

Step-by-step explanation:

Given:

Angle, θ = 35°

Vertical distance, Δx = 6 m

Diameter, d = 1.9 cm

= 0.019 m

A.

When the water leaves the sprinkler, it does so at a projectile motion.

Therefore,

Using equation of motion,

(t × Vox) = 2Vo²(sin θ × cos θ)/g

= Δx = 2Vo²(sin 35 × cos 35)/g

Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)

= 62.57

Vo = 7.91 m/s

B.

Area of sprinkler, As = πD²/4

Diameter, D = 3 × 10^-3 m

As = π × (3 × 10^-3)²/4

= 7.069 × 10^-6 m²

V_ = volume rate of the sprinkler

= area, As × velocity, Vo

= (7.069 × 10^-6) × 7.91

= 5.59 × 10^-5 m³/s

Remember,

1 m³ = 1000 liters

= 5.59 × 10^-5 m³/s × 1000 liters/1 m³

= 5.59 × 10^-2 liters/s

= 0.0559 liters/s.

For the 4 sprinklers,

The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559

= 0.224 liters/s

C.

Area of 1.9 cm pipe, Ap = πD²/4

= π × (0.019)²/4

= 2.84 × 10^-4 m²

Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s

= 2.24 × 10^-4 m³/s

Velocity of the water, Vw = volumetric flowrate/area

= 2.24 × 10^-4/2.84 × 10^-4

= 0.787 m/s

7 0
3 years ago
Read 2 more answers
Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste
lidiya [134]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

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3 years ago
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Using the 10% value to find 20% of 300
neonofarm [45]
20% * 300 = 2 * (10% * 300) = 2*30 = 60
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3 years ago
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