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3 years ago

11

Can someone solve this please?

1 answer:

fiasKO [112]3 years ago

8 0

Answer: x = 1/2

Step-by-step explanation:

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Get y by itself <br> 5x+y=10

almond37 [142]

We can add in any order. So 5x+y is the same as y+5x

5x+y = 10
y+5x = 10
y+5x-5x = 10-5x ... subtract 5x from both sides
y+0x = -5x+10
y+0 = -5x+10
y = -5x+10

Answer: y = -5x+10

Note: this is in y=mx+b form where m = -5 is the slope and b = 10 is the y intercept

8 0

3 years ago

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Solve for x in both problems

Gnom [1K]

Answer:

x=3 and x=5

Step-by-step explanation:

its half of the bottom number

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3 years ago

Which is the correct way to write 850.05 in words on a check

zysi [14]

Eight hundred fifty and five cents

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3 years ago

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In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

<u>1. Start with the first 11 inspected packages:</u>

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

<u>2. The 12th package</u>

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

<u>3. Finally</u>

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

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3 years ago

Write the solution to the given inequality in interval notation.

Effectus [21]

Is the last one
(-Infinity, -3] U (-1, positive infinity)

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