Really need help plss

A weight-loss clinic advertised that last month its patients, on average,8lb lost . Assuming that average refers to the mean, wh

Ipatiy [6.2K]

Answer:

Given:

A weight-loss clinic advertised that last month its patients, on average, lost 8 lbs

Average refers to mean

a. Last month all of their patients lost at least 8 lbs.

Answer: Mean or average is a measure of central tendency or in other terms refers to the central value of the dataset. If the value of mean/average is 8lb, it signifies that there are some patients who have lost weights more than 8 lbs. So, this statement is False

b. Last month at least one of their patients lost exactly 8 lbs.

Answer: It is a possibility that all the patients have lost their weights greater than or less than 8 lbs. Since the mean is computed by summing up all the values (weight loss in this case) by the total number of observations, it is possible that there can be no patient who has lost exactly 8lbs. So, this statement is False.

c. Two months ago some of their patients lost at least 8 lbs.

Answer: It is clearly mentioned that last month its patients, on average, lost 8 lbs. No information is given about the weight loss of patients before the last month. So, there is not sufficient data to check whether this statement is correct or not.

d. Last month at least one of their patients lost less than 3 lbs.

Answer: It is a possibility that all the patients have lost their weights greater than 3 lbs. Since the mean is computed by summing up all the values (weight loss in this case) by the total number of observations, it is possible that there can be no patient who has lost less than 3lbs. So, this statement is False.

e. Last month, the number of their clients who lost less than 8 lbs was equal to the number of their clients who lost more than 8 lbs.

Answer: Mean is a measure of central tendency which is computed by the diving sum of all observations with the number of all observations (the sum of observations lesser than the mean is compensated by the sum of observations greater than the mean) and for the mean equal to 8lbs, there should be an equal number of people who lost less than 8 lbs and the number of people who lost more than 8 lbs. Hence, this statement is True.

Answer e.Last month, the number of their clients who lost less than 8 lbs was equal to the number of their clients who lost more than 8 lbs is True

7 0

2 years ago

Read 2 more answers

Simplify the expression

aliina [53]

The answer is definately A

4 0

3 years ago

Read 2 more answers

Colin walked a distance of 15 miles in 6 hours. <br> Work out out

butalik [34]

Answer:

Colin walked 2.5 miles every hour

Step-by-step explanation:

The answer is 2.5 because 15 divided by 6 is 2.5

P.S Can I have brainliest?

8 0

2 years ago

Using differential calculus, maximize the volume of a box made of cardboard (top is open) as shown in Figure A. 15, subject to t

lana [24]

The maximum volume of the box is 40√(10/27) cu in.

Here we see that volume is to be maximized

The surface area of the box is 40 sq in

Since the top lid is open, the surface area will be

lb + 2lh + 2bh = 40

Now, the length is equal to the breadth.

Let them be x in

Hence,

x² + 2xh + 2xh = 40

or, 4xh = 40 - x²

or, h = 10/x - x/4

Let f(x) = volume of the box

= lbh

Hence,

f(x) = x²(10/x - x/4)

= 10x - x³/4

differentiating with respect to x and equating it to 0 gives us

f'(x) = 10 - 3x²/4 = 0

or, 3x²/4 = 10

or, x² = 40/3

Hence x will be equal to 2√(10/3)

Now to check whether this value of x will give us the max volume, we will find

f"(2√(10/3))

f"(x) = -3x/2

hence,

f"(2√(10/3)) = -3√(10/3)

Since the above value is negative, volume is maximum for x = 2√(10/3)

Hence volume

= 10 X 2√(10/3) - [2√(10/3)]³/4

= 2√(10/3) [10 - 10/3]

= 2√(10/3) X 20/3

= 40√(10/27) cu in

To learn more about Maximization visit

brainly.com/question/14682292

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Complete Question

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3 0

1 year ago

For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe

Rus_ich [418]

To solve the inequality $\frac{x^2+x+3}{2x^2+x-6}\ge \:0$

$\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:$

The numerator $x^2+x+3$ is not factorizable.

so factor the denominator $2x^2+x-6$ :

$2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\$