Lists bout 4 packs of socks for 36$ how many packs of socks can she buy for 54$ Find unit rate to solve this

Pls help gsoehehsjbsgshsjbsfeoanevshksnevehwknsbdgsknwbsgdindvegshsn

Kryger [21]

Answer:

a) 5.9

b) 4.2

Step-by-step explanation:

Plot the data points on the number line (see attached).

Part (a)

$\begin{aligned}\textsf{Distance between Ben and Kloee} & = \sf \left| Ben - Kloee \right|\\& = \sf \left|-2.65-3\frac{1}{4}\right|\\& = \sf \left|-2.65-3.25\right|\\& = \sf \left|-5.9\right|\\& = \sf 5.9\end{aligned}$

Part (b)

$\begin{aligned}\textsf{Distance between Middle school and Football field} & = \sf \left| Middle\:school-Football\:field\right|\\& = \sf \left| 1.5-5.7\right|\\& = \sf \left| -4.2 \right|\\& = \sf 4.2\end{aligned}$

4 0

2 years ago

Help please :). I hate trigonometry

elena-s [515]

Answer: You shouldn't hate trig.

Step-by-step explanation:

SinΘ= opposite/hypotenuse

SinΘ=y/x

Sin(16) = 9/x

x = 9/Sin(16)

x= 32.7 cm

7 0

3 years ago

How do I prove that 2 cot 2x = cot x-tan x?

BlackZzzverrR [31]

We will start off working on the right hand side.
cot x - tan x 
= [cos x / sin x] - [sin x / cos x] 
= [(cos x)^ 2 - (sin x)^2] / [sin x cos x] 

This is where it gets a bit tougher if you do not have your formula list with you. 
(cos x)^ 2 - (sin x)^2 = cos(2x) 
sin 2x = 2 sin x cos x 

Note that by arranging the second formula, we will have sin x cos x = (1/2) sin 2x 

Hence, we will get: 
[(cos x)^ 2 - (sin x)^2] / [sin x cos x] 
= [cos 2x] / (1/2)[sin 2x] 
= 2[cos 2x] / [sin 2x] 
= 2cot 2x

7 0

4 years ago

Write the equation of the circle. A with a center A(-4,-3) and radius 5

tiny-mole [99]

Answer:

(x+4)^2 + (y+3)^2 = 5^2

Step-by-step explanation:

The equation of a circle is

(x-h)^2 + (y-k)^2 = r^2

now substitute the known values

(x+4)^2 + (y+3)^2 = 5^2

and theres your equation

4 0

4 years ago

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