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4 years ago

Enter the new function after the given transformation

1 answer:

6 0

H₁(x)=(x+8-3)³-5=(x+5)³-5 is the function h(x) 3 units right
h₂(x)=(x+8)³-5+3=(x+8)³-2 is the funtion h(x) 3 units up

h₃(x)=(x+8-3)³-5+3=(x+5)³-2 is the funtion h(x) 3 unist right and 3 units up.

answer: the new funtion will be: h(x)=(x+5)³-2

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Solve the system of equations. 5x y = 9 3x 2y = 4 (−2, 5) (1, 4) (2, −1) (4, −4)

kobusy [5.1K]

The solution of the system of equations that is given is (2,-1).

Given a system of equations are 5x+y=9 and 3x+2y=4.

A system of linear equations (or linear system) is a collection of one or more linear equations involving the same variables. A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied.

The given equations are

5x+y=9 ......(1)

3x+2y=4 ......(2)

Here, the substitution method is used to solve the system of equations.

Find the value of y from equation (1) by subtracting 5x from both sides.

5x+y-5x=9-5x

y=9-5x

To find the value of x substitute the value of y in equation (2).

3x+2(9-5x)=4

Apply the distributive property a(b+c)=ab+ac as

3x+2×9-2×5x=4

3x+18-10x=4

Combine the like terms on the left side as

-7x+18=4

Subtract 18 from both sides and get

-7x+18-18=4-18

-7x=-14

Divide both sides by -7 and get

(-7x)÷(-7)=(-14)÷(-7)

x=2

Substitute the value of x in equation (1) and get

5(2)+y=9

10+y=9

Subtract 10 from both sides

10+y-10=9-10

y=-1

Hence, the solution of system of equations 5x+y=9 and 3x+2y=4 is (2,-1).

Learn more about system of equations from here brainly.com/question/13729904

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6 0

2 years ago

A movie streaming service charges it’s customers $15 a month. Martina has $98 saved up. Will she have any money left over if she

lukranit [14]

Answer:

98 : 15 = 6 R 8

She can pay for six months 6*15 = $90 and she will have $8 left.

6 0

3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

Inessa05 [86]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319