Can you help me please

An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba

madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

X = a person is overweight

Y = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688$

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

$P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312$

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0

3 years ago

8) 8x + 6 14x - 2 A) 11 C) 8 B) -9 D) -7

Travka [436]

Answer:

C) 8

Step-by-step explanation:

Consecutive angles are equal to 180. So 22x+4+180. Solve that to get x=8. Hope I helped and post more questions :)

6 0

3 years ago

8) through: (-3, -2), perp. to y = x – 1 A) y=-5x – 1 B) y=-4x – 5 C) y=-x – 5 D) y=-5x – 4

nexus9112 [7]

Answer:

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

Solution:

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1$

We know that, product of slopes of perpendicular lines is -1.

So, 1 $\times$ slope of perpendicular line = -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

8 0

3 years ago

eryn and seven of her freinds go out to lunch the bill comes out 143.35 and they leave a 15% tip how much does each freind contr

babunello [35]

Answer:

$ 23.55

Step-by-step explanation:

143.35*.15=21.5025 + 143.35 = $164.8525/7 = $ 23.5503571429

7 0

3 years ago

A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag fo

defon

Answer:

a) The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

b) $P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}, a \leq x \leq b$

And using this function we got:

$P(X$

c) $P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$

Step-by-step explanation:

For this case we define the random variable X=quantity put in each bag , and we know that the distribution for X is given by:

$X \sim Unif (a= 10, b =12)$

Part a

The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

Part b

For this case we want this probability:

$P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}, a \leq x \leq b$

And using this function we got:

$P(X$

Part c

For this case we want this probability:

$P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$

3 0

3 years ago

Can you help me please ​

Can you help me please