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Andreas93 [3]
3 years ago
11

What is the slope of a line parallel to the line whose equation is y = 5x + 4?

Mathematics
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

5

Step-by-step explanation:

y=mx+b

m=slope

in your equation y=5x+4, m=5

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Consider the following information related to a set of quantitative sample data:
svlad2 [7]

Answer:

(a) There are outliers

(b) x and x >62

Step-by-step explanation:

Given

\sigma = 14.92

\bar x = 22.0

q_0 = -24

q_1 = 14.5

q_2 = 24.5

q_3 = 33.5

q_4 = 64

Solving (a): Check for outliers

This is calculated using:

Lower = Q_1 - (1.5 * IQR) --- lower bound of outlier

Upper = Q_3 +(1.5 * IQR) --- upper bound of outlier

Where

IQR = Q_3 - Q_1

So, we have:

IQR = 33.5 - 14.5

IQR = 19

The lower bound of outlier becomes

Lower = Q_1 - (1.5 * IQR)

Lower = 14.5 - (1.5 * 19)

Lower = 14.5 - 28.5

Lower = -14

The upper bound of outlier becomes

Upper = Q_3 +(1.5 * IQR)

Upper = 33.5 + 1.5 * 19

Upper = 33.5 + 28.5

Upper = 62

So, we have:

-14 \le x \le 62 --- the range without outlier

Given that:

q_0 = -24  --- This represents the lowest data

q_4 = 64   --- This represents the highest data

-24 and 64 are out of range of -14 \le x \le 62.

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

x and x >62

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3 years ago
The radius of a circle is 5m, what is the diameter​
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Answer:

10

Step-by-step explanation:

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8 0
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A fisherman uses a spring scale to weigh a tilapia fish. He records the fish weight as a kilograms and notices that the spring s
belka [17]

Answer:

k = \frac{980a}{b}

Step-by-step explanation:

Fisherman noticed a stretch in the spring = 'b' centimetres

Weight of the fish = a kilograms

If force applied on a spring scale makes a stretch in the spring then Hook's law for the force applied is,

F = kΔx

Where k = spring constant

Δx = stretch in the spring

F = weight applied

F = mg

Here 'm' = mass of the fish

g = gravitational constant

F = a(9.8)

  = 9.8a

Δx = b centimetres = 0.01b meters

Therefore, 9.8a = k(0.01b)

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k = \frac{980a}{b}

Therefore, spring constant of the spring will be determined by the expression, k = \frac{980a}{b}

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