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3 years ago

Which equation has the solution y=2?

Mathematics

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

Step-by-step explanation:

solve A

2y - 3 = 19

2y = 22

y = 11

solve B

3y + 2 = 8

3y = 6

solve C

4y - 4 = -4

4y = 0

y = 0

solve D

5y + 1 = 10

5y = 9

y = 9/5

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Solve for x in the equation x^2+4x-4 = 8.

ollegr [7]

Answer:

Its the first choice.

Step-by-step explanation:

x^2+4x-4 = 8

x^2 + 4x - 4 - 8 = 0

x^2 + 4x - 12 = 0

(x + 6)(x - 2) = 0

x = -6, 2.

8 0

3 years ago

Find the inner product for (7, 2) * (0, -2) and state whether the vectors are perpendicular.

Andrews [41]

Answer: A

Step-by-step explanation:

To find the inner product of two vectors (a,b) and (c,d) you would use the equation (a * c) + (b * d)

So for (7,2) and (0,-2) the inner product would be

(7 * 0) + (2 * -2)

= 4

The vectors are only perpendicular when the inner product is equal to 0. Since it is equal to -4 in this case, the vectors are not perpendicular.

A -4; no

3 0

3 years ago

Read 2 more answers

Which function is the inverse of ? f(x)=50,000(0.8)^x

IrinaK [193]

Answer:

y=log(x)−log(50000)log(0.8)

Explanation:

write as : y=50000(0.8)x

Taking logs:

log(y)=log(50000)+log(.(0.8)x.)

But log(.(0.8)x.) is the same as xlog(0.8)

Thus
x=log(y)−log(50000)log(0.8)

Now swap the x'x and the y's giving:

y=log(x)−log(50000)log(0.8)

my teacher helped a little bit

4 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Is (1, 2) a solution of 6x-y > 3?

nikdorinn [45]

Answer:

yes

Step-by-step explanation:

6(1)-2>3

6-2>3

4>3

3 0

3 years ago