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2 years ago

6

Small boxes contain DVDs, and large boxes contain one gaming machine. Three boxes of gaming machines and a box of DVDs weigh 48

pounds. Three boxes of gaming machines and five boxes of DVDs weigh 72 pounds. How much does each box weigh?

1 answer:

Pani-rosa [81]2 years ago

7 0

I don’t know so sorry maybe ask someone else but just go back to the passage and see if you can find it

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What number has the absolute value of 16

german

The Answer is -16

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2 years ago

Find the amount of time. I=$237.90, P=$1525, r=2.6%<br><br> think I=PRT

egoroff_w [7]

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3 years ago

How many terms in 9a+11p

Naddik [55]

Answer:

2 terms

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2 years ago

Read 2 more answers

Helen got a 7% reduction in the price of pair jeans that are normally $32. What is the approximate percent decrease in the price

Colt1911 [192]

A percentage is a way to describe a part of a whole. The approximate percent decrease in the price is 21.87%, thus, the correct option is B.

<h3>What are Percentages?</h3>

A percentage is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25 which is equal to 25%.

To convert a fraction to a percentage, convert the fraction to decimal form and then multiply by 100 with the '%' symbol.

Given the reduction in the price is $7, while the original cost is $32. Therefore, the approximate percent decrease in the price is,

Percentage decrease = 7/32 × 100%

= 21.87%

Hence, the approximate percent decrease in the price is 21.87%, thus, the correct option is B.

The correct question is:

Helen got a $7 reduction in the price of a pair of jeans that are normally $32. What is the approximate percent decrease in the price?

The percent decrease is approximately 15%.

The percent decrease is approximately 21%.

The percent decrease is approximately 43%.

The percent decrease is approximately 78%

Learn more about Percentages:

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3 0

1 year ago

The half-life of cesium-137 is 30 years. Suppose we have a

VMariaS [17]

(a) If $y(t)$ is the mass (in mg) remaining after $t$ years, then $y(t) = y(0) e^{kt}=100e^{kt}.$

$y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}$

(b) $y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}$

(c)
$100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}$

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3 years ago