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3 years ago

13

Given an infinite population that does not meet the conditions of a normal distribution, samples taken at random will also not b

e normally distributed regardless of sample size.
True
False

2 answers:

KatRina [158]3 years ago

7 0

Answer:

False

Step-by-step explanation:

The central limit theorem says that if a sufficient large sample is taken from a population with finite variance, the mean of all the samples will be the population mean.

IN other words, irrespective of the original distribution from where samples were drawn, the mean of all samples for large sample sizes will follow a normal distribution.

Hence the given statement is false.

For sample sizes large, we can expect samples will have mean that follows a normal distribution.

AVprozaik [17]3 years ago

6 0

Answer:

100% false

Reading the answer at this very moment

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When a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a ra

Sergeu [11.5K]

Answer:

$r=\sqrt{10}\text{ m}$

Step-by-step explanation:

We have been given that when a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a rate of 0.7 meters per hour. We are asked to find the approximate radius of tank in meters.

We will use volume of cylinder formula to solve our given problem as:

$V=\pi r^2h$ , where,

r = Radius,

h = Height of cylinder.

Since the level of water in the tank rises at a rate of 0.7 meters per hour, so height of cylinder would be $h = 0.7$ meters at $V=22\text{ m}^3$ .

Upon substituting these values in above formula, we will get:

$22\text{ m}^3=\frac{22}{7}\cdot r^2(0.7\text{ m})$

$22\text{ m}^3=\frac{22}{10}\cdot r^2\text{ m}$

$\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}=r^2$

$r^2=\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}$

$r^2=10\text{ m}^2$

Now, we will take positive square root of both sides as radius cannot be negative.

$\sqrt{r^2}=\sqrt{10\text{ m}^2}$

$r=\sqrt{10}\text{ m}$

Therefore, radius of tank would be approximately square root of 10 m.

5 0

2 years ago

Karen has $1.70 in coins. Karen has 8 coins, all of which are quarters or dimes.

kumpel [21]

x+y=8 and 25x+10y=170 are the linear equations.

x+y≤8 and 25x+10y≤170 are the inequalities.

Step-by-step explanation:

Given,

Worth of coins = $1.70 = 1.70*100 = 170 cents

Number of coins = 8

1 quarter = 25 cents

1 dime = 10 cents

Let,

x represent the number of quarters

y represent the number of dimes

1. Write an equation to represent the amount of coins Karen has.

x+y = 8

2.Write an equation to represent the value of the coins Karen has.

25x+10y=170

x+y=8 and 25x+10y=170 are the linear equations.

For inequalities, the amount cannot increase number of coins and worth but it can be less, therefore,

x+y≤8

25x+10y≤170

x+y≤8 and 25x+10y≤170 are the inequalities.

Keywords: linear equations, addition

Learn more about linear equations at:

#LearnwithBrainly

4 0

3 years ago

Find the midpoint of the segment having endpoints (0 ,1/8) and (-4/5 ,0)

kvasek [131]

Answer:

Mid point of the given end points = $(\frac{-2}{5},\frac{1}{16} )$

Step-by-step explanation:

Given the end points of the line segment are :

(0, $\frac{1}{8}$ ) & ( $\frac{-4}{5}$ ,0)

We will use the mid point formula when two points are: (x₁,y₁) & (x₂,y₂)

mid point = $(\frac{x_{1} +x_{2} }{2},\frac{y_{1} +y_{2} }{2} )$

now put the value of x₁ = 0

x₂ = $\frac{-4}{5}$

y₁ = $\frac{1}{8}$

y₂ = 0

mid point = $(\frac{0-\frac{4}{5} }{2},\frac{\frac{1}{8}+0 }{2})$

= $(\frac{-2}{5},\frac{1}{16} )$

That's the final answer.

3 0

3 years ago

What is the equation of the line that passes through (-3,-1 ) and has a slope of 3/5

vfiekz [6]

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (-3, -1) and the slope m = 3/5. Substitute:

$y-(-1)=\dfrac{3}{5}(x-(-3))\\\\y+1=\dfrac{3}{5}(x+3)\qquad\text{use distributive property}\\\\y+1=\dfrac{3}{5}x+\dfrac{9}{5}\qqua\text{subtract 1 from both sides}\\\\y=\dfrac{3}{5}x+\dfrac{4}{5}\qquad\text{multiply both sides by 5}\\\\5y=3x+4\qquad\text{subtract 3x from both sides}\\\\-3x+5y=4\qquad\text{change the signs}\\\\3x-5y=-4$

Answer:

point-slope form: y + 1 = 3/5(x + 3)

slope-intercept form: y = 3/5x + 4/5

standard form: 3x - 5y = -4

7 0

3 years ago

GIVING BRAINLIEST! A rectangular garden has a width of 10 feet and a length of 14 feet. A cement walkway is added around the out

Serjik [45]

Given:

Width of a garden = 10 feet

Length of the garden = 14 feet

Area of the garden and the walkway together = 396 square feet.

To find:

The width of the walkway.

Solution:

A cement walkway is added around the outside of the garden.

Let x be the width of the walkway.

The width of the garden with walkway = 10+2x

The length of the garden with walkway = 14+2x

Area of a rectangle is

$Area=length\times width$

Area of garden and the walkway together is

$Area=(14+2x)\times (10+2x)$

$396=140+28x+20x+4x^2$

$396=140+48x+4x^2$

$396=4(35+12x+x^2)$

Divide both sides by 4.

$99=35+12x+x^2$

$0=35-99+12x+x^2$

$0=-64+12x+x^2$

Splitting the middle term, we get

$x^2+16x-4x-64=0$

$x(x+16)-4(x+16)=0$

$(x+16)(x-4)=0$

Using zero product property, we get

$(x+16)=0\text{ and }(x-4)=0$

$x=-16\text{ and }x=4$

Width of walkway cannot be negative. So, x=4.

Therefore, the width of the walkways is 4 feet.

6 0

2 years ago

Read 2 more answers