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Wittaler [7]
3 years ago
12

Round 50.468 to the nearest thousandtha)50.000b)51.000c)51.400d)51.500​

Mathematics
1 answer:
Olenka [21]3 years ago
6 0
Ok it should be B if I’m rounding that up right
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I am a 3-digit decimal. The sum of my digits is nine. One of my decimal digits is 0. I have a 3 in my tenths place. Who am I?
dezoksy [38]
.360 or .306
Because 3 is in the tenths place, there is a 0, and 3+6+0= 3+6=9 the sum of all the digits. And it’s a three digit decimal.
6 0
3 years ago
10 mL = ____________ L ( 1L = 1000 ml
DENIUS [597]

Solution:

<u>Note that:</u>

  • 1 L = 1000 ml

<u>Using the clue above, let's solve each problem.</u>

  • 10 mL        =        10/1,000 L           =           0.01 L
  • 1.2 L          =     1.2 x 1,000 mL        =       1,200 mL
  • 3,500 mL =     3,500/1,000 L        =           3.5 L
  • 4 L            =       4 x 1,000 mL        =       4,000 mL
  • 230 mL    =        230/1,000 L         =           0.23 L
  • 6.21 L       =     6.21 x 1,000 mL      =       6,210 mL

Hoped this helped!

4 0
3 years ago
The distance s (in feet) covered by a car t sec after starting is given by the following function. s = −t3 + 11t2 + 20t (0 ≤ t ≤
VladimirAG [237]

8.94

Answer:

Step-by-step explanation:

6 0
4 years ago
Can someone please explain how to go the answers for this
Effectus [21]

Answer:

A=3, B=4, C=-5

Step-by-step explanation:

So A represents the amplitude, which is the distance between the midline (C), the middle of the graph, and the maximum (or minimum). B is the distance of the period of the graph. In this case, we see that the midline is C=-5 because that is the halfway point of the sine function. We also see that the period distance is B=5-1=4, so our period would be 2π/4 or π/2. Our amplitude would be A=3 because |a|=|-5-(-2)|=|-5+2|=|-3|=3. Observe the graph to see this visually.

7 0
3 years ago
In a class of 50 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose
Tems11 [23]

Answer:  3

Step-by-step explanation:

Let E be the event of that student pierces ear and N be the event of that student pierces nose.

Given: n(E\cup N=50)

n(E)=46\\\\n(N)=7

For any two event A and B, we have

n(A\cup B)=n(A)+n(B)-n(A\cap B)

Similarly , n(E\cup N)=n(E)+n(N)-n(E\cap N)

50=46+7-n(E\cap N)\\\\\Rightarrow\ n(E\cap N)=53-50=3

Hence, 3 students have piercings both on their ears and their noses.

7 0
3 years ago
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