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3 years ago

Continuous variable expected to show a statistical relationship with dependent variables are called ____

Mathematics

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

Instant Variables

Step-by-step explanation:

A continuous variable is a variable defined as one which can receive an infinite set of values.

This continuous variable has two types namely instant variables and ratio variables.

Instant variables are those in which the distance between each category is equal or static. Meanwhile, ratio variables are those in which the ratio between the given scores provide information in regards to the relationship between the responses/dependent variables.

Thus, the one that corresponds to our question are instant variables.

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The rate of change is 1/4 and the y intercept is -4

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3 years ago

Variables x, y, and z are integers such that xyz > 0. If xy=30, x/z=2/3, and z=9, what is the value of y?

Xelga [282]

<h3>Answer: 5</h3>

Explanation:

Plug z = 9 into the second equation and solve for x.

x/z = 2/3

x/9 = 2/3

x = 9*(2/3)

x = 18/3

x = 6

This leads to

xy = 30

y = 30/x

y = 30/6

y = 5

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3 years ago

Use the diagram to help you solve the equation 4x - 12 = 16

KatRina [158]

Answer: X=7

Step-by-step explanation:

Add 12 to both sides.

Divide both sides by 4.

Then you'll get 7 for X

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3 years ago

A thermometer is taken from a room where the temperature is 20◦C to the

lord [1]

Answer:

a) $T(2) = 8.265$ : at time t = 2 minutes the temperature will be 8.265 degress

b) $6 = T(3.55)$ : the temperature will be 6 degrees at time t = 3.55 minutes.

Step-by-step explanation:

When dealing with temperature changes, it's best to work with Newton's Law of Cooling.

$T(t) = T_s + Ce^{kt}$

here:

$T(t)$ : the temperature in the room.

$T_s$ : ambient (or outdoor) temperature (that always remains constant, in our case: $T_s = 5$ )

$C\,\text{and}\,k$ : are constants

Our conditions are provided:

1) $T(0) = 20$

2) $T(1) = 12$

using the first condition

$T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15$

using the second condition:

$T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}$

we can use our calculated value of C to find k

$e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621$

Finally we can put these constants back in the main equation:

$T(t) = T_s + Ce^{kt}$

$T(t) = 5 + 15e^{-0.7621t}$ or $T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}$

a) Reading after one more minute?

so it's asking:

T(2) = ?

$T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267$

Hence, after one more minute the temperature of the room will be 8.267 degrees

b) When will it be 6 degrees?

T(t) = 6?

$6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\$

Hence at t = 3.55 minutes the temperature of the room will be 6 degrees.

8 0

3 years ago

Using the number line below find the

bonufazy [111]

Answer:

0.3

Step-by-step explanation:

To find this probability, we need to know the distance from point H to P and the distance from point L to K, and then the probability of a random point in HP being inside LK will be the division of the distance from L to K by the distance from H to P:

H to P: distance = 15 - 5 = 10

L to K: distance = 10 - 7 = 3

Probability: LK/HP = 3/10 = 0.3

8 0

3 years ago