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3 years ago

What percent of 80 is 34

1 answer:

7 0

To find the percentage, you can multiply (34/80) by (100/1) to obtain the percentage. (This applies for all fractions)

In this case, you will get 3400/80 = 42.5%

Hope this helps!

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she has 2 1/2 cups of peanuts. 3 1/4 cups of rainsins and 2 2/3 cups of banana chips. how many cups of trail mix can she make

andriy [413]

Answer:

8 cups of trail mix

Step-by-step explanation:

2 1/2 + 3 1/4 + 2 2/3= 8 5/12

6 0

3 years ago

Read 2 more answers

Let's say the probability of you winning a game of checkers and a game of chess against your friend is 20%. The probability of y

GREYUIT [131]

The probability of winning a game of checkers against your friend is 67%.

<h3>What is the probability of winning a game of checkers against my friend?</h3>

Probability can be described as the process of determining the chances of an event happening. The chances that an event would occur has a value that lies between 0 and 1. A value of 0 is given when the event does not occur and a value of 1 if the event occurs.

The probability of winning a game of checkers = probability of winning both games / probability of winning a game of chess

20% / 30% = 67%

To learn more about probability, please check: brainly.com/question/26321175

8 0

2 years ago

What is the decimal equivalent of 16/3

puteri [66]

Answer:
5.33333

How to:
16/3

7 0

3 years ago

Read 2 more answers

You are investing $1500 at 5.2% compounded annually. How much money will you have in 12 years?

erica [24]

Answer:

$2756.01 to the nearest cent.

Step-by-step explanation:

A = P(1 + r)^t where A is the amount after t years, P = amount invested, r = rate as a decimal fraction.

So here, we have:

A = 1500(1 + 0.052)^12

= 1500 * 1.837337

= $2756.01

8 0

2 years ago

What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$

7 0

3 years ago