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3 years ago

A 32-ounce package of a snack mix is 24 percent pretzels. How many ounces are pretzels?

2 answers:

8 0

Answer: The correct answer is Choice D.

If 24 percent of the snack mix is pretzel, we simply need to multiply the total number of ounces by 0.24.

0.24 x 32 = 7.68

7.68 ounces of the snack mix will be pretzels.

sammy [17]3 years ago

6 0

Answer:

The Answer Is D.)

Step-by-step explanation:

I took the test!!

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In a doctor's waiting room, there are 14 seats in a row. Eight people are waiting to be seated. There is someone with a very bad

diamong [38]

Answer: 12,972,960

Step-by-step explanation: There are 14 chairs and 8 people to be seated. But among the 8. three will be seated together:

So 5 people and (3) could be considered as 6 entities:

Since the order matters, we have to use permutation:

¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆ x 3!

2,162,160 x 6 = 12,972,960 ways.

Another way to solve this problem is as follow:

5 + (3) people are considered (for the time being) as 6 entities:

The 1st has a choice among 14 ways

The 2nd has a choice among 13 ways

The 3rd has a choice among 12 ways

The 4th has a choice among 11 ways

The 5th has a choice among 10 ways

The 6th has a choice among 9ways

So far there are 14x13x12x11x10x9 = 2,162,160 ways

But the 3 (that formed one group) could seat among themselves in 3!

or 6 ways:

Total number of permutation = 2,162,160 x 6 = 12,972,960

5 0

3 years ago

Pls help! I need to know if this is right! 50 points!

Sedaia [141]

Answer: B) (x-1)2+(y+1)2= 25 radius = 5 miles

Step-by-step explanation:

In order to get this answer you see that you are adding a negative to x in the first part then multiplying it by 2. Second, you add y+1 then multiplying it by 2. You get the radius of 25 which equals 5 miles. Look at the graph to also check your answers to the problem.

4 0

3 years ago

Read 2 more answers

In Example 3, the track has 6 lanes that are each 1 meter in width. a. What is the outer perimeter of the track? Round your answ

son4ous [18]

Answer:

Incomplete question, check attachment for the necessary diagram

Step-by-step explanation:

Note in the attachment,

We have two identical straight line of lenght

L1 = L2 = 84.39m

We also have two identical semicircle or radius 36.5m to the first track lane

But this is not the radius of the circle, the radius of the circle will now be 36.5 plus the 6 track lane and we are told that one track lane is 1m, then, the track lane is 6m

So, radius = 36.5+6

r = 42.5m

Then, we need to calculate the perimeter of the semicircle using the formula of perimeter of a circle and dividing by2

P = 2πr/2

P =πr

P = 22/7 × 42.5

P = 133.57 m

Then, the arc 1 is equal to arc 2 which is equal to 133.57 m

A1 = A2 = 133.57 m

Now we have all the dimensions,

Then, the perimeter can be calculated by adding the length of the sides

The perimeter of the field = Lenght of the two straight lines plus the length of the two semicircle arc

P = L1 + L2 + A1 + A2

P = 84.39 + 84.39 + 133.57 + 133.57

P =435.923 m

So, to the nearest meter

P ≈ 436m

The perimeter of the track is 436m

5 0

3 years ago

The graph shows a proportional relationship between the number of kilometers traveled by a bicycle (y) and the number of minutes

vredina [299]

Answer:

3.10 i think i saw in quiz i hope you do a great job and ace the test

Step-by-step explanation:

6 0

2 years ago

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago