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2 years ago

Which is an example of a statistical question?

Mathematics

2 answers:

Phantasy [73]2 years ago

6 0

<h2>Answer:</h2>

The example of a statistical question is:

Question 4: How old are the students in math class?

<h2>Step-by-step explanation:</h2>

Statistical question--

A statistical question is one which is based over the data which is sampled or collected such that there is a variability in the data.

Here the first three questions i.e.

Question 1 How old am I?

Question 2 How old is the teacher?

Question 3 How old will I be after my next birthday?

are not based over the sample these are based over a individual and not over the data or population. Hence, they are invalid.

The statistical question is:

D) Question: 4

Vikki [24]2 years ago

5 0

Hey there, Damonealy!

The correct answer should be D since you are talking about multiple amount of data. As you can see, question number 1-3 are talking about the age of a single person but question D is talking about the ages of multiple students which can be used in a statistic.

Thank you for using Brainly.
See you soon!

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Solve the system of equations.

guajiro [1.7K]

Answer:

Option b. (x = 3, y = 20, z = -14)

Step-by-step explanation:

Given:

2x + 2y + 3z = 4

5x + 3y + 5z = 5

3x + 4y + 6z = 5

Solve using Cramer’s rule

∴ $\left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] =\left[\begin{array}{ccc}4\\5\\5\end{array}\right]$

∴A = $\left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] = -1$

Ax = $\left[\begin{array}{ccc}4&2&3\\5&3&5\\5&4&6\end{array}\right] = -3$

Ay = $\left[\begin{array}{ccc}2&4&3\\5&5&5\\3&5&6\end{array}\right] =-20\\$

Az = $\left[\begin{array}{ccc}2&2&4\\5&3&5\\3&4&5\end{array}\right] = 14$

∴ x = Ax/A = -3/-1 = 3

y = Ay/A = -20/-1 = 20

z = Az/A = 14/-1 = -14

So, the answer is option b. (x = 3, y = 20, z = -14)

4 0

3 years ago

Consider the logarithms base 5. For each logarithmic expression below, either calculate the value of the expression or explain w

Tom [10]

Answer:

log₅(3125) = 5

Step-by-step explanation:

Given:

log₅(3125)

Now,

using the property of log function that

logₐ(b) = $\frac{\log(b)}{\log(a)}$

thus,

Therefore, applying the above property, we get

⇒ $\frac{\log(3125)}{\log(5)}$ (here log = log base 10)

now,

3125 = 5⁵

thus,

⇒ $\frac{\log(5^5)}{\log(5)}$

Now,

we know from the properties of log function that

log(aᵇ) = b × log(a)

therefore applying the above property we get

⇒ $\frac{5\log(5)}{\log(5)}$

⇒ 5

Hence,

log₅(3125) = 5

8 0

2 years ago

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

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