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3 years ago

What is the slope between (20,5) and (17,11)

Mathematics

1 answer:

Andre45 [30]3 years ago

8 0

Answer:

Step-by-step explanation:

Slope: (y₂ - y₁) / (x₂ - x₁)

Putting in the values we have:

(11 - 5) / (20 - 17) =

6 / 3 = 2

Thus, the slope is 2.

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What is the equation of the line that passes through the point (-3,-8) and has a slope of 4?

coldgirl [10]

Step-by-step explanation:

the easiest approach with a given point and the slope of the line is the point-slope form :

y - y1 = a(x - x1)

where "a" is the slope, and (x1, y1) is a point on the line.

so, we get

y - -8 = 4(x - -3)

y + 8 = 4(x + 3)

if we need the slope-intercept form

y = ax + b

we now simplify the point-slope form

y + 8 = 4x + 4×3 = 4x + 12

y = 4x + 4

3 0

2 years ago

Steven earns extra money babysitting. He charges $31.25 for 5 hours and $50 for 8 hours. Write an equation to represent the rela

hichkok12 [17]

I’m not sure but this seems right

Y=6.25X

4 0

3 years ago

Help me solve this?

LiRa [457]

Answer:

Step-by-step explanation:

5(-5)-2=-27

5(-2)-2= -12

5(1)-2=3

5(2)-2=8

5(3)-2=13

x g(x)

-5 -27

-2 -12

1 3

2 8

3 13

5 0

3 years ago

Read 2 more answers

You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 15 km/hr and your

Viefleur [7K]

Answer:

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin (2πt))

b) the foot have to be 3.32 rev/sec faster

Step-by-step explanation:

Given that:

the speed of the bike = 15 km/hr = 15 × 1000/3600 (m/sec) = 4.167 m/sec

radius of the circle when the foot moves = 20 cm = 0.2 m

radius of the circle above the ground = 30 cm = 0.3 m

Let assume that:

x(t) should represent the vector along the horizontal moment

y(t) should be the vector along the vertical moment

The initial component will be ( 0, 0.3)

We know that the radius of the circle is given as 0.2 m, So the vector of the circle can be written as (0.2 cos t , 0.2 sin t )

Also, the foot makes one revolution in a second, definitely the frequency of the revolution = 1 and the vector for the circle is ( 0.2 cos (2πt), -0.2 sin (2πt)), due to the fact that the foot moves clockwise.

Thus, adding all the component together ; we have:

(x(t), y(t)) = (0,0.3)+(4.167 t , 0)+(0.2 cos (2πt), -0.2 sin (2πt))

(x(t), y(t)) = (4.167 t + 0.2 cos (2πt), 0.3 - 0.2 sin (2πt))

Hence; the parametric equations are:

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin (2πt))

The linear speed of rotation is :

15km/hr = 15 × 100, 000/3600 (cm/sec)

= 416.7 cm/sec

The rotational frequency is :

= 416.7/2πr

= 416.7/2(3.14 × 20)

= 3.32 rev/sec

Hence, the foot have to be 3.32 rev/sec faster in rotating if an observer standing at the side of the road sees the light moving backward.

4 0

3 years ago

Read 2 more answers

Which of the following is not a property of a chi-square distribution?

laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

$\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2$

Where N₁ , N₂m .... $N_k$ are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0

3 years ago