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1 year ago

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in one version of a trail mix, there are 3 cups of peanuts mixed with 2 cups of raisins in another version of trail mix, there a

re 4.5 cups of peanuts mixed with 3 cups of raisins. are the ratios equivalent for the two mixes?

1 answer:

vlada-n [284]1 year ago

8 0

Based on the ratios of the peanuts mixed with raisins in the versions of the trail mix, the ratios are equivalent for both mixes.

<h3>What are the ratios?</h3>

The first ratio is:

3 : 2

When taken to the simplest terms it is:

3/2 : 2/2

1.5 : 1

The second ratio is:

4.5 : 3

4.5/3 : 3/3

1.5 : 1

The ratios of the peanut and raisins mix are therefore equivalent.

Find out more on equivalent ratios at brainly.com/question/2328454

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Mr. Hayden's Bible class collected $11

ki77a [65]

This is the only way I can think of this mathematically. This took a good 20 minutes for me. You need to guess and check. Like try 17 quarters, 50 nickels, and 37 dimes etc. until you get 11.00. My numbers are mathematically correct but we both know you cant have 18.909 quarters so. lol

Answer:

Quarters: 18.909

Nickels: 49.818

Dimes: 37.818

Somewhere around those numbers.

Step-by-step explanation:

$11.00 total in coins

Nickel = $0.05

Dime = $0.10

Quarter = $0.25

(n*0.05) + (d*0.10) + (q*0.25)

Solve the equation for q.

$q=\frac{1}{2} d$

(n*0.05) + (d*0.10) +( $\frac{1}{2} d$ *0.25)

Simplify the expression.

9d + 2n = 440

d - n = -12

Multiply both sides of the equation by -9.

-9d + 9n = 108

9d + 2n = 440

Sum the equations vertically to eliminate at least one variable.

11n = 548

$n=\frac{548}{11}$

Substitute the given value of n into the equation d - n = -12

$d-\frac{548}{11} =-12$

$d=\frac{416}{11}$

Substitute the given value of n into the equation $q=\frac{1}{2} d$

$q=\frac{1}{2} *\frac{416}{11}$

$q=\frac{208}{11}$

Finally

$d=\frac{416}{11}$ d = 37.818

$n=\frac{548}{11}$ n = 49.818

$q=\frac{208}{11}$ q = 18.909

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Answer:

Step-by-step explanation:

We know that between 1 to 10 there are 5 even and 5 odd numbers.

We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards

Let´s check all this combinations

Case 1: When all 4 numbers are even:

We are going to take 4 of the 5 even numbers in the box so we have

$5C4=5$

Case 2: When all 4 numbers are odd:

We are going to take 4 of the 5 odd numbers in the box, so we have

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Case 3: When 2 are even and 2 are odd:

We are giong to take 2 from 5 even and odd cards in the box so we have

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Remember that we obtain the probability from

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So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:

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Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
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