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Vikki [24]
3 years ago
13

How many lines of symmetry does the letter “A” have?

Mathematics
2 answers:
kykrilka [37]3 years ago
7 0

Answer:

1 line

The rest of the letters, A, B, C, D, and E all have only 1 line of symmetry. Notice that the A has a vertical line of symmetry, while the B, C, D, and E have a horizontal line of symmetry.

Eva8 [605]3 years ago
3 0
One Line of Symmetry
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Hunter-Best [27]

Step-by-step explanation:

WX is parallel to YZ and WZ is also parallel to XY

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3 years ago
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faltersainse [42]

Step-by-step explanation:

1 .f(x)=x^{2}-x+1

   f(-1)= (-1)-(-1)+1

   f(-1)= 1+1+1

      = 3

3.f(x)=x^{2}-x+1

 f( 1)=(1)-1+1

  f( 1)=1-1+1

        =1

5. f(x)=x^{2}-x+1

    f(3)=(3)^{2}-3+1

         = 9-3+1

          =7

2. g(x) = 5 - 3x

 g( -8)=5-3(-8)

g( -8)=5+24

        =29

4.g(x) = 5 - 3x

g(5)=5-3(5)

 g(5)=5-15

       =-10

6.g(x) = 5 - 3x

   g(-3)=5-3(-3)

 g(-3)=5+9

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3 0
2 years ago
URGENT PLEASE HELP! What is the slope of the line <br> y = -2x - 4?
Tju [1.3M]

Answer:

-2 is the slope

Step-by-step explanation:

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3 years ago
What's the answer to. 2472 ÷ 24
vladimir2022 [97]

Answer:

Hello!!! Princess Sakura here ^^

Step-by-step explanation:

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5 0
3 years ago
A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o
Brilliant_brown [7]

Answer:

The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.

Step-by-step explanation:

Let <em>X</em> = number of students who have laptops.

The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.

A random sample of <em>n</em> = 30 students is selected.

The event of a student having a laptop is independent of the other students.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean and standard deviation of a binomial random variable <em>X</em> are:

\mu=np\\\sigma=\sqrt{np(1-p)}

Compute the mean of the random variable <em>X</em> as follows:

\mu=np=30\times0.37=1.11

The mean of the random variable <em>X</em> is 1.11.

Compute the standard deviation of the random variable <em>X</em> as follows:

\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836

The standard deviation of the random variable <em>X</em> is 0.836.

5 0
3 years ago
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