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Alexxandr [17]
3 years ago
10

I need help please with this math problem

Mathematics
1 answer:
Arada [10]3 years ago
8 0
When the dilation is centered on the origin, you can multiply each of the individual coordinates by the scale factor.

All the coordinates are ±5 and the scale factor is 1/5. Multiplying those gives ±1. The only selection with all coordinates being ±1 is
.. Selection A.

You can check to see if the signs agree in detail. (They do.)
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Who ever gets this answer right will get a brainlest
n200080 [17]

Answer:

The answer is a Line. (2nd option)

Step-by-step explanation:

3 0
3 years ago
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What is a simplest way to find out x
konstantin123 [22]

Answer:

You should start with a simple equation. And remember- practice makes perfect.

The basic algebraic equation involves simple addition or subtraction with one unknown quantity, such as 2 + x = 7.

1.  Subtract 2 from both sides: 2 - 2 + x = 7 - 2.

2. Simplify the equation by doing the math: 2-2+x=7-2=0+x=5, or x = 5.

3. Check your work by substituting the answer, 5, into the equation for x.

The correct answer is x=5.




5 0
3 years ago
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Suppose a normal distribution has a mean of 79 and a standard deviation of 7. What is P(x ≥ 72)?
Zina [86]

Approximately 65% of the distribution lies within one standard deviation of the mean, which is to say,

P(72 ≤ x ≤ 86) ≈ 0.65

Normal distributions are symmetric, so the percentage of values one standard deviation below the mean is equal to the percentage of values one standard deviation above the mean.

P(72 ≤ x ≤ 79) = P(79 ≤ x ≤ 86)

but since the sum of these make up P(72 ≤ x ≤ 86), we find

P(72 ≤ x ≤ 79) ≈ 0.65/2 = 0.325

Also due to symmetry, exactly half of the distribution lies to either side of the mean; namely,

P(x ≥ 79) = 0.5

It follows that

P(x ≥ 72) = P(72 ≤ x ≤ 79) + P(79 ≤ x)

P(x ≥ 72) = 0.325 + 0.5

P(x ≥ 72) = 0.825 ≈ 0.84

8 0
3 years ago
Determine the number of real solutions for each system of equations.
Lisa [10]

Answer:

1+1=2

Step-by-step explanation:

7 0
4 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
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