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laiz [17]
4 years ago
9

Solve 41-9+(8×2.3)-15+(2.1×4)

Mathematics
1 answer:
dsp734 years ago
6 0
(8x2.3)= (18.4)
(2.1x4)= (8.4)
Now you should have:
41-9+18.4-15+8.4
9+18= 27
15+8.4= 23.4
Should have:
41-27-23.4= -9.4
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I need help with this
sleet_krkn [62]
First lets invert the divisor and chane to a multuplication:-
       x + 5                     2x^2 - 7x - 4
-----------------        *   --------------------
2x^2 + 3x + 1           3x^2 + 18x + 15

Now we can factor  the 3 trinomials:-
      x + 5                   (2x + 1)(x - 4)
 --------------         *   ------------------
(2x + 1)(x + 1)        3(x + 5)(x + 1)

Now we simplify because (2x + 1) and (x + 5) are common to top and bottom:-
            1             *        x - 4
=  -------------           ------------
       x + 1              3( x + 1)

=       x - 4
    --------------
     3(x + 1^2


5 0
3 years ago
Write the standard form of the equation. through (2-4) slope=-1
svetoff [14.1K]
(Y+4)=-1m (X-2)
Hopes this helps 
6 0
3 years ago
I really need help , can anybody please help me???
marta [7]

Answer:no, there are always other ways. LIke this y> x.

Step-by-step explanation:

I use schoology too

4 0
3 years ago
Select the property that allows the left member of the equation to be changed to the right member.
lapo4ka [179]
That would be the symmetric property
3 0
4 years ago
Read 2 more answers
Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to
Novosadov [1.4K]

Answer:

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

X \sim N(\mu = 150000, \sigma =2300

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

7 0
4 years ago
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