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NikAS [45]
2 years ago
13

Simplify : 48÷6 [-52 ÷2 {4 - 3 (2 - 15÷3)}]​

Mathematics
1 answer:
garri49 [273]2 years ago
8 0

Answer:

86

Step-by-step explanation:

48÷6 [-52 ÷2 {4 - 3 (2 - 15÷3)}] \\ 8[-52 ÷2 {4 - 3 (2 - 15÷3)}] \\ 8[-52 ÷2{1 (2 - 15÷3)}] \\ 8[ - 2.5{ (2 - 15÷3)}] \\ 8[ - 2.5{ ( - 13÷3)}] \\ 8[ - 2.5{ (  - 4.3)}] \\ 8[10.75 { }] \\  = 86

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Subtract 2xy from 6xy<br>​
dmitriy555 [2]

Answer:

4xy

Step-by-step explanation:

We need to subtract 2xy from 6xy, So that is,

=> 6xy - 2xy

=> 4xy

6 0
2 years ago
Read 2 more answers
21×9=9×21 what property is used
kvv77 [185]
Hello There =)

Commutative property is used. The commutative property <span>states that two numbers can be multiplied in either order.</span>


5 0
3 years ago
Keith as 11/12 hours to play, he has already played 1/4. which is the best estimate of the fractional part of an hour he has lef
____ [38]

Answer:

The best estimate is greater than 1/2 but less than 3/4

Step-by-step explanation:

If you multiply 1/4 by 3/3 (which is also 1), you can change the denominator without changing the value.  So 1/4 is equal to 3/12.

Since keith has 11/12 hours to play, and he has already played 3/12 hours, subtract 3/12 from 11/12 to get 8/12 hours. This is how much time he has left to play.  

If you simplify 8/12 hours, you get 2/3 hours.

So the best estimate would be: greater than 1/2 but less than 3/4.

4 0
3 years ago
While completing a race, Will spent 14 minutes walking. If his ratio of time walking to jogging was 7 : 6, how many minutes did
worty [1.4K]

Answer:

12 minutes

Step-by-step explanation:

\frac{7}{6} =\frac{14}{x}

7x=84

x=12

So, he spent 12 minutes jogging.

8 0
3 years ago
The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric
Papessa [141]
Let X denote the number of minutes spent brushing his teeth, and let \mu be the mean and \sigma the standard deviation for this distribution.

\mathbb P(X

The z-score corresponding to this probability is approximately z=-0.2533, which means

\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1

Next, (note the sign change)

\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98

The corresponding z-score is approximately z=2.0538, so you have

\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2

Solving the two equations for \mu and \sigma, you'll find that the mean is approximately \mu=1.1098 and the standard deviation is approximately \sigma=0.4334.
7 0
3 years ago
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