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Andreyy89
2 years ago
14

Using a scale of 4cm to 1 unit on both axis, what does that mean?​

Mathematics
1 answer:
I am Lyosha [343]2 years ago
5 0

Answer:

It means that for every 4cm on an axis, it counts as 1 unit.

So for instance, you have a 24cm axis to work with. You would have a total of 6 units on the axis, because each unit is 4cm.

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Classify the polygons as regular or irregular, and concave or convex. HELP ASAP!! TWO PICTURES!!
mylen [45]

Answer:

First figure is regular and convex.

Second figure is irregular and convex.

Step-by-step explanation:

Here given 2 polygons we have to classify these whether these are convex, concave, regular and irregular.

Regular Polygon is the polygon which have all sides and all angles equal.

Convex Polygon is the polygon in which measure of all interior angles less than 180° and if measure of each of interior angle greater than 180° then it is concave.

In the first figure, all sides are equal therefore and each of interior angle of polygon is less than 180°. Hence, the polygon is regular and convex.

In the second figure, all sides are not equal  and each of interior angle of polygon is less than 180°. Hence, the polygon is irregular and convex.

6 0
2 years ago
Solve for x.<br><br> −(−2−5x)+(−2)=18<br><br> x=−90<br> x=90<br> x=185<br> x=−185
Tomtit [17]
Uhm, did you, maybe, type in the problem wrong? The answer to this equation is: 18/5 or 3.6, lol, no worries, though!
5 0
3 years ago
Read 2 more answers
Can someone pls help me ?
Firlakuza [10]
-4 is the correct answer. -24= -4/6
8 0
3 years ago
If f(x)=5x^4+3, then what is the remainder when f(x) is divided by x-1?
KatRina [158]
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This is actually fairly simple. If you take the x-1 and set it equal to 0 you get 1. And if you plug that value into the x you get 8, which is the remainder. This is called the remainder value theorem.
3 0
3 years ago
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Given the probability distributions shown to the​ right, complete the following parts.
Elan Coil [88]

Answer:

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

Variance = Var(X) = 10.46 - 3.02² = 1.3396

Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

7 0
2 years ago
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