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ra1l [238]
3 years ago
7

WRITE THE TERM,FILL IN THE BLANK

Mathematics
2 answers:
xeze [42]3 years ago
6 0

Answer:

It is simplify. It is what you do when like terms are combined in order to get your answer.

Hope this helps!

lianna [129]3 years ago
5 0

The answer you’re looking for is: simplify.

When you rewrite an expression, we are simplifying it. We first use the distributive property through any groupings symbols, then evaluate terms we exponent that are given. Lastly, we combine any like terms, weather they are constants or variables.

Hope this helps! :)

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in January, Dawn earns $9.25 allowance. She earns 3 times as much in February. If during March, she earns $5.75 more than she di
AURORKA [14]
Jan: 9.25
Feb: 9.25 x 3 = 27.75
March: 27.75 + 5.75 = 33.50

Dawn earns $33.50
8 0
3 years ago
Read 2 more answers
What is (3xy^2)(-6xy^4)
krok68 [10]

Answer:

-18x^2y^6

Step-by-step explanation:

we have

(3xy^2)(-6xy^4)

Applying property of exponents

x^{m} x^{n}=x^{m+n}

(3xy^2)(-6xy^4)=(3)(-6)(x^{1+1})(y^{2+4})=(-18)(x^{2})(y^{6})=-18x^2y^6

5 0
3 years ago
(4.04 LC)
sukhopar [10]
The answer will be B
4 0
3 years ago
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A company hires five people for the same job for one week. The amount that each person is paid
dangina [55]

Answer:

20

Step-by-step explanation:

4 0
3 years ago
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A trimming operation at a local manufacturer produces rods whose length conforms to unifrom distribution with a minimum of 18cm
MariettaO [177]

Answer:

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

Step-by-step explanation:

Given: uniform distribution with min and max values of 18 and 33 respectively.

To find : probability density for upper cumulative frequency i.e 25.8 at least means x\geq25.8  upto 33 cm i.e the maximum limit of function.

Solution:

we have by definition , of uniform distribution

we get , probability density function defines as :

<em>F(x,a,b)= </em>\frac{1}{b-a}<em>   </em>a\leq x\leq b

            =1/(33-18)=1/15=0.0667.

this is probability density function.

here the x=25.8 , a=18 and b=33

for lower cumulative frequency it defines as ;

P(x,a,b)=\frac{x-a}{b-a} =25.8-18/33-18=0.52

for upper cumulative frequency it defines as ;

Q(x,a,b)=b-x/b-a=33-25.8/33-18=0.48

here at least 25.8 cm probability means it should be greater than a value(18cm) hence it is provided by the upper cumulative frequency

i.e. Q(x,a,b)=0.48

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

7 0
3 years ago
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