Solve the following using substitution

Mathematics

2 answers:

zubka84 [21]2 years ago

7 0

The image is not opening for me

Hoochie [10]2 years ago

5 0

Send a different documentStep-by-step explanation: ill do it if you send it

You might be interested in

Plz guys help me In 2 please

Doss [256]

Answer:

Step-by-step explanation:

OMG!!!!! why don't you just SAY YOU WANT a BLEEPING REFLECTION!!!

I had to search through a slew of your questions to get clues as to what to do.

To reflect over the x axis, just negate the y term and leave the x term as it is.

M becomes (2, -1)

N becomes (-3, -1)

P becomes (-1, -4)

7 0

2 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

Work varies directly with the

irinina [24]

Answer:

yes

Step-by-step explanation:

6 0

2 years ago

Julian spends 85 min on HW each day. If 20% of the time is spent on math HW, how many minutes a day is Julian spending on math H

allsm [11]

The answer is 17 minutes