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Scorpion4ik [409]
3 years ago
14

What 2 numbers should be used to find 231 - 198 using the 10 method for subtracting quickly in your head?

Mathematics
1 answer:
Airida [17]3 years ago
4 0

Answer: "Count 2 and then count 31 more."

Step-by-step explanation:

We have the equation:

231 - 198

Now, the negative number is kinda ugly, so i will write it as:

200 - 2 = 198

Is a lot easier work with 200 and 2, than with 198.

Then the equation is now:

231 - (200 - 2)

And the left number we also have a "200", so it can be written as:

200 + 31 = 231

As this is a sum, we can ignore the parentheses.

200 + 31 - 200 + 2

31 + 2

Then the correct option is:

"Count 2 and then count 31 more."

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Crazy question if you answer all of the steps you get 15 points
viktelen [127]
AC = 32

2x + 6x + 8 = 32   (plug in into equation)

8x + 8 = 32 (simplify)

8x + 8 (- 8) = 32 (-8) (subtraction of equality,)
(what you do to one side, you do to the other)
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5 0
3 years ago
13 Answer:
Marizza181 [45]

Step-by-step explanation:

13.

132 miles in 3 h = 132miles/3hours = 132/3 miles/h =

= 44 miles/h

so,

how long for 110 miles ?

speed = distance/time

time = distance/speed = 110 miles / 44 miles/h =

= 110/44 h = 55/22 h =

= 5/2 h = 2.5 h

14.

60% = 60/100 = 6/10 = 3/5

15.

300% = 300/100 = 3

16.

116 2/3% = (348 + 2)/3% = 350/3% = 350/3/100 =

= 350/300 = 1 50/300 = 1 1/6

17.

19% = 19/100

18.

3.8% = 3.8/100 = 38/1000 = 19/500

19.

166 2/3% = (498 + 2)/3% = 500/3% = 500/3/100 =

= 500/300 = 5/3 = 1 2/3

20.

4/5 = 0.8 = 80%

21.

7/4 = 1.75 = 175%

22.

1/3 = 0.3333333... ≈ 33.33%

23.

2 = 200/100 = 200%

24.

0.4 = 40%

25.

0.375 = 37.5/100 = 37.5%

26.

80% of 60 = 60×80/100 = 60×4/5 = 48

27.

24% of 65 = 65×24/100 = 65×6/25 = 13×6/5 =

= 15.6

28.

115% of 138 = 138×115/100 = 138×23/20 =

= 69×23/10 = 158.7

29.

18.3% of 74 = 74×18.3/100 = 74×183/1000 =

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30.

6.5% of 115 = 115×6.5/100 = 115×65/1000 =

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31.

0.75% of 93 = 93×0.75/100 = 93×75/10000 =

= 93×3/400 = 0.6975 ≈ 0.70

4 0
2 years ago
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igor_vitrenko [27]
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8 0
3 years ago
The lifetime of two light bulbs are modeled as independent and exponential random variables X and Y, with parameters lambda and
Scorpion4ik [409]

Answer:

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Step-by-step explanation:

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\bf f(x)=\lambda e^{-\lambda x}\;(x\geq0)

The PDF of Y is

\bf g(x)=\mu e^{-\mu x}\;(x\geq0)

The means of X and Y are respectively,

\bf \displaystyle\frac{1}{\lambda}\;,\displaystyle\frac{1}{\mu}

so we can see that the larger the parameter, the smaller the mean. Hence the PDF of Z = min(X, Y) is an exponential with the largest parameter of the two.

Therefore, the PDF of Z is

\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)

7 0
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Answer:

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Step-by-step explanation:

7 0
4 years ago
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