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OLga [1]
3 years ago
7

Which situation can be modeled using a linear function?

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

the distance covered by a car at a constant speed

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Find the missing value in each figure below. What does “y” equal?
finlep [7]

Answer:

Step-by-step explanation:

The perpendicular is equal to 6. That's because the left triangle's missing angle is 180 - 45 -90 = 45

The angle in the right triangle is given as 52.

The cos(52) = adjacent side (which we just found to be 6) / y

Multiply both sides by y

y cos(52) = 6

cos(52) = 0.6157

Divide by sides by cos(52)

y = 6 / cos(52)

y = 6 / 0.6157

y = 9.76

8 0
3 years ago
What is the inequality shown in the graph
rodikova [14]

It must be ≤ instead of ≥.

y ≤ - 3·x + 4

3 0
3 years ago
April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the
zaharov [31]

Answer:

The maximum height of the arrow is 125 feet.

Step-by-step explanation:

The pathway of the arrow can be represented by the equation,

h = -16t^2 +80t + 25 .....(1)

Where h is height in in feet and t is time in seconds.

It is required to find the maximum height of the arrow. For maximum height, \dfrac{dh}{dt}=0.

So,

\dfrac{d(-16t^2 +80t + 25)}{dt}=0\\\\-32t+80=0\\\\t=\dfrac{80}{32}\\\\t=2.5\ s

Put t = 2.5 s in equation (1). So,

h = -16(2.5)^2 +80(2.5) + 25\\\\h=125\ \text{feet}

So, the maximum height of the arrow is 125 feet.

6 0
3 years ago
Lines AB and CD are parallel. If the measure of angle
BARSIC [14]

Answer:

Angles W, Z, M, P are 111°

Angles Y, X, O, N are 69°

3 0
3 years ago
Suppose we roll a fair six-sided die 20 times and draw ten cards from a standard 52-card deck. Let X be the number of "6"s rolle
Lera25 [3.4K]

Answer:

a) Expected value = 6.406

Variance = 4.905

Standard deviation = 2.45

b) The probability is 0.08547

Step-by-step explanation:

a) Let's suppose that:

X₁ = number of 6´s

X₂ = number of Jack, Queen, King or Aces

The mean of X₁ is:

MeanX₁ = n * p = 20 * (1/6) = 3.33

The variance of X₁ is:

Var-X_{1} =np(1-p)=3.33(1-(1/6))=2.775

The mean of X₂ is:

MeanX₂ = 10 * (16/52) = 3.076

The variance of X₂ is:

Var-X_{2} =3.076(1-(16/52))=2.13

The expect value of X is:

Xexp = MeanX₁ + MeanX₂ = 3.33 + 3.076 = 6.406

The variance of X is:

VarX = VarX₁ + VarX₂ = 2.775 + 2.13 = 4.905

The standard deviation is:

Xdevi = 4.905/2 = 2.45

b) The probability of drawing at least five six out of 20 rolls is equal to:

∑(1/6)ˣ(5/6)²⁰⁻ˣ = 0.231 with x = 5

The probability of at least 4 Jack, Queen, Kings or Aces is:

∑(16/52)ˣ(1-(16/52))¹⁰⁻ˣ = 0.37 with x = 4

The probability of given event is equal to:

P = 0.231 * 0.37 = 0.08547

5 0
3 years ago
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