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3 years ago

I need help with number 9

Mathematics

2 answers:

Westkost [7]3 years ago

7 0

Answer:

A) A dilation followed by a reflection.

Step-by-step explanation:

I went over this in class the other day. (:

Vilka [71]3 years ago

3 0

Answer:

Step-by-step explanation:

Any time you use a dilation, you destroy congruency (unless the dilation factor is one). The diluted image will be bigger or smaller than the original.

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PLEASE HELP ASAP!!!!!!

Anvisha [2.4K]

Answer:

1/5

Step-by-step explanation:

3 0

3 years ago

Pls show me proof, thank you

Karo-lina-s [1.5K]

The y has to be negative but the x positive. So that leaves only (3,-4)

4 0

3 years ago

If you walk 5/12 in 1/2 hour how dar will you walk in 1 hour

riadik2000 [5.3K]

Answer:

10/12 or 5/6

Step-by-step explanation:

1/2 + 1/2 = 1

5/12 + 5/12 = 10/12 or 5/6

3 0

3 years ago

Read 2 more answers

The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

Anit [1.1K]

Answer: The correct option is A.

Step-by-step explanation: We are given a polynomial which is a sum of other 2 polynomials.

We are given the resultant polynomial which is : $8d^5-3c^3d^2+5c^2d^3-4cd^4+9$

One of the polynomial which are added up is : $2d^5-c^3d^2+8cd^4+1$

Let the other polynomial be 'x'

According to the question:

$8d^5-3c^3d^2+5c^2d^3-4cd^4+9=x+(2d^5-c^3d^2+8cd^4+1)$

$x=8d^5-3c^3d^2+5c^2d^3-4cd^4+9-(2d^5-c^3d^2+8cd^4+1)$

Solving the like terms in above equation we get:

$x=(8d^5-2d^5)+(-3c^3d^2+c^3d^2)+(5c^2d^3)+(-4cd^4-8cd^4)+(9-1)$

$x=6d^5-2c^3d^2+5c^2d^3-12cd^4+8$

Hence, the correct option is A.

5 0

3 years ago

Read 2 more answers

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293

4 0

3 years ago