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3 years ago

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Which is a solution of the inequality?p + 4 – 2(p – 10) > 0

2 answers:

boyakko [2]3 years ago

5 0

Hi Laila

p+4 -2(p-10)>0

The first thing you need to do is simplify the both sides

So that will be

p-2p + 4+ 20 > 0

-p + 24 > 0

-p > 0-24

-p > -24

Now since we can not have the negative sign in front of P, we need to do something. What we need to do is divide both sides by -1

-p/-1 > -24/-1

p < 24

Remember to change the symbol.

I hope that's help and if you have any further questions please let me know :)

san4es73 [151]3 years ago

3 0

Answer:

I got 24 as being WRONG! ITS NOT 24!! the other answer is wrong! its dumb that it got verified, verified by who!

This is the REAL ANSWER: i don't know, just pick 26 or something. i looked and looked all over the internet and no one had any other answer other than 24! so yeah..... just guess and if you get it right please leave a comment to say the real answer. I already got rid of 1 for you. GOOD LUCK! 33.33% chance!

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A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

krek1111 [17]

Answer:

The probability that the first two electric toothbrushes sold are defective is 0.016.

Step-by-step explanation:

The probability of an event, say <em>E </em>occurring is:

$P (E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let <em>X</em> = number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is, <em>n</em> = 20.

Number of defective electric toothbrushes is, <em>x</em> = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

$=\frac{3}{190}\\$

$=0.01579\\\approx0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

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Step-by-step explanation:

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