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2 years ago

The graph below is the solution for which set of inequalities?

Mathematics

1 answer:

kipiarov [429]2 years ago

6 0

Answer:

R u in ms Francis class??

Step-by-step explanation:

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-1 1/5 + 3/4 I’m bad with fractions and don’t know how to do these it’s due tomorrow

Dovator [93]

-1 1/5 + 3/4 = -0.45

1/5 is 0.2 in decimal form, and 3/4 is 0.75:

-1.2 + 0.75 = -0.45

Hope this helps!

8 0

3 years ago

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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

The length of a rectangle with area 38 in2 depends on its width. Write the formula that describes the length of the rectangle as

Ostrovityanka [42]

Answer:

L = 38/W

Step-by-step explanation:

The area of a rectangle is computed by multiplying length by width. For length L and width W in inches, your rectangle has area ...

A = LW

38 = LW

Then the length can be found by dividing the equation by W:

L = 38/W

8 0

3 years ago

Which function has no horizontal asymptote? F (x) = StartFraction 2 x minus 1 Over 3 x squared EndFraction f (x) = StartFraction

yanalaym [24]

Answer:

$f(x)=\frac{2x^2}{3x-1}$

Step-by-step explanation:

The function $f(x)=\frac{2x-1}{3x}$ has a horizontal asymptote which is $y=\frac{2}{3}$

The function $f(x)=\frac{x-1}{3x}$ has a horizontal asymptote which is $y=\frac{1}{3}$

The function $f(x)=\frac{2x^2}{3x-1}$ has no horizontal asymptote because the numerator has a degree which is higher than the degree of the denominator,

The function $f(x)=\frac{3x^2}{x^2-1}$ has a horizontal asymptote which is $y=3$

3 0

3 years ago

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Simplify: −3(4x + 5)(2x + 4)

san4es73 [151]

Sorry if I’m wrong!!

3 0

3 years ago

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