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Anton [14]
3 years ago
12

What is this answerSimplify to create an equivalent expression. \qquad{8k-5(-5k+3)}8k−5(−5k+3)

Mathematics
2 answers:
Licemer1 [7]3 years ago
6 0

Answer:

33k - 15

Step-by-step explanation:

Given expression,

\qquad{8k-5(-5k+3)}

By the distributive property,

=8k-5(-5k) - 5(3)

=8k+25k-15

By combining like terms,

=33k-15

∵ Further simplification is not possible,

Hence, the required simplified form of the given expression is,

33k-15

marissa [1.9K]3 years ago
3 0

Answer:

33k-15

Step-by-step explanation:

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Hiya! can anyone help me with these? it would be greatly appreciated!
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Answer:

K would be -6 and w would be 9

5 0
3 years ago
Read 2 more answers
8x+y=300
Bond [772]
D. The elevator takes 37.5 seconds to move down to the lobby.


8x+y=300

y= 300-8x
y= 300 - 8 (37.5)

y = 300-300
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5 0
3 years ago
How many blocks fill this prism?
zaharov [31]
Answer : 72 blocks

To find the answer, you have to find the volume of the prism. To find the volume you multiply the length, width, and height.

6 x 3 x 4 = 72

This gives you 72 blocks.
3 0
3 years ago
The least common multiple of two numbers is 60. the prime factorization of one number is 3 times 5. what is the prime factorizat
gulaghasi [49]
The first number would be 3×5=15. 
Another number could have different values. it could be
4=2×2
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7 0
3 years ago
Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
ziro4ka [17]

Answer:

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.75241,0.06281)  

Where \mu=5.75241 and \sigma=0.06281

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

For the case when z =-2 we can do this:

-2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 -2*0.06281 =5.6121

And for the other case when Z=2 we have:

2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 +2*0.06281 =5.8886

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

5 0
3 years ago
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