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patriot [66]
3 years ago
6

-3/8 as a decimal hurrry

Mathematics
2 answers:
AveGali [126]3 years ago
5 0

Answer:

- 3/8 as a decimal is -0.375

Step-by-step explanation:

luda_lava [24]3 years ago
5 0

Answer:

-0.375

Step-by-step explanation:

Non-negitive would be 0.375.

Each 1/8 is 0.125

0.125x3=0.375

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When answered please show step by step
Allushta [10]
How am I supposed to answer that if I can't get the graph to you?
3 0
3 years ago
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a s
Korvikt [17]

Answer:

On the graphing calculator, use the function normCdf, where

  • lower bound = -9999
  • upper bound = 210
  • mean = 250
  • standard deviation = 46

It will result in normCdf(-9999,210,250,46) ≈ 0.192269 or 19.2269%

6 0
3 years ago
I need help what is 3.45 x (-2.8)<br> please show steps
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3 0
3 years ago
A shark weighs 405 kg and 68 grams a second shark weighs 324 kg and 75 g how much more does the first shark weigh in grams than
mariarad [96]

For this case we have to by definition:

1 kg equals 1000 grams

Shark 1:

405 kg and 68 grams

405 kg * \frac {1000 g} {1kg} = 405,000 grams

Thus, shark 1 weighs 405,068 grams.

Shark 2:

324 kg and 75 grams

324 kg * \frac {1000 g} {1 kg} = 324,000 grams

Thus, shark 2 weighs 324,075 grams.

Subtracting we have:

405.068 grams-324.075 grams = 80.993 grams

Thus, shark 1 weighs 80,993 grams more than the second.

Answer:

Shark 1 weighs 80,993 grams more than the second.

8 0
3 years ago
How many gallons of a 80% antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antif
konstantin123 [22]

440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

<em><u>Solution:</u></em>

Let "x" be the gallons of 80 % antifreeze added

Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze

Final mixture is x + 80

Therefore, we can frame a equation as:

"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze

Thus, we get,

x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %

x \times \frac{80}{100} + 80 \times \frac{15}{100} = (x+80) \times \frac{70}{100}\\\\0.8x + 80 \times 0.15 = (x+80) \times 0.7\\\\0.8x+12 = 0.7x+56\\\\0.8x-0.7x=56-12\\\\0.1x = 44\\\\x = \frac{44}{0.1}\\\\x = 440

Thus 440 gallons of 80 % antifreeze solution must be mixed

4 0
3 years ago
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