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3 years ago

Solve the solution. How many solutions does it have? 4(2x-3)=2(4x-6)

Mathematics

1 answer:

sdas [7]3 years ago

7 0

Answer:

No solutions

Step-by-step explanation:

4(2x-3)=2(4x-6)

$4(2x-3)=2 \times 2(2x-3) \\ \\ 4(2x-3)=4(2x-3)$

Same components are present on both sides of equal sign, therefore NO SOLUTIONS are possible.

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Jeff has 4 red pens and 2 blue pens in his backpack. He also has 1 yellow highlighter and 4 green highlighters in his

Romashka-Z-Leto [24]

Answer:

$\frac{1}{15}$

Step-by-step explanation:

In probability,

"AND" means "multiplication"

"OR" means "ADDITION"

We want probability of Blue Pen and Yellow Highlighter. So we find individual probabilities and "MULTIPLY" them.

Let P(b) be probability of blue pen

Let P(y) be probability of yellow highlighter

We know probability of event is that event divided by total number of that

So,

There are 4 + 2 = 6 pens

Blue pen = 2

Hence,

P(b) = 2/6 = 1/3

And,

There are 5 highlighters

Yellow highlighter = 1

Hence,

P(y) = 1/5

Now,

P(b and y) = 1/3 * 1/5 = 1/15

Hence,

probability of grabbing blue pen and a yellow highlighter is $\frac{1}{15}$

3 0

3 years ago

Read 2 more answers

48 is what percent of 60?

Mariulka [41]

48 is 80% of 60

Divide 48 over 60:
$\frac{48}{60} = 0.8$

Convert your decimal to a percentage:
$0.8 \times 100 = 80$

5 0

3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago

In which quadrant is the ordered pair ( 7, -8 ) located? *

Paul [167]

Answer:

i think IV

Step-by-step explanation:

because of the picture

4 0

3 years ago

Read 2 more answers

QUESTION 7.

Elan Coil [88]

The probability that exactly 4 of the selected adults believe in reincarnation is 5.184%, and the probability that all of the selected adults believe in reincarnation is 7.776%.

Given that based on a poll, 60% of adults believe in reincarnation, to determine, assuming that 5 adults are randomly selected, what is the probability that exactly 4 of the selected adults believe in reincarnation, and what is the probability that all of the selected adults believe in reincarnation, the following calculations must be performed:

0.6 x 0.6 x 0.6 x 0.6 x 0.4 = X
0.36 x 0.36 x 0.4 = X
0.1296 x 0.4 = X
0.05184 = X
0.05184 x 100 = 5.184

0.6 x 0.6 x 0.6 x 0.6 x 0.6 = X
0.36 x 0.36 x 0.6 = X
0.1296 x 0.6 = X
0.07776 = X
0.07776 x 100 = 7.776

Therefore, the probability that exactly 4 of the selected adults believe in reincarnation is 5.184%, and the probability that all of the selected adults believe in reincarnation is 7.776%.

Learn more in brainly.com/question/795909

5 0

2 years ago