Ask question

Ask question

All categories

3 years ago

6

Hello i need help solving the following inequality X (X + 3) ⩽ X (2X + 5) thanks

1 answer:

igor_vitrenko [27]3 years ago

3 0

Answer:

-2 ⩽ x

Step-by-step explanation:

x(x + 3) ⩽ x (2x + 5) divide both sides of tge inequality with x

x + 3 ⩽ 2x + 5 subtract x from both sides

3 ⩽ x + 5

-2 ⩽ x

You might be interested in

A quadratic equation of the form 0 = ax2 + bx + c has a discriminant value of –16. How many real number solutions does the equat

vredina [299]

$\bf \qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ \stackrel{-16}{negative}&\textit{no solution}~~\checkmark \end{cases}$

6 0

3 years ago

Read 2 more answers

It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha

stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

0.54% of the calls receive a busy signal, hence p = 0.0054.

A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

$P(X \geq 5) = 1 - P(X < 5)$

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009$

$P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062$

$P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218$

$P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513$

$P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903$

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295$

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

2 years ago

Please answer this i dont understnad

serg [7]

The answer would be 6fg^3+5f^2g^2+f^3g-7

7 0

4 years ago

48 cookies 25% increase

9966 [12]

Answer:

60

Step-by-step explanation:

48 multiplied by 25% = 48 multiplied by 0.25 = 12

48 plus the increase = 60

5 0

3 years ago

Patsy was playing softball. When she got up to bat, she struck out once, was tagged out twice, walked three times and had hit 5

icang [17]

4×5=20 is the answer

8 0

3 years ago