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1 year ago

I need help like rn plzzzz No linksss

Mathematics

2 answers:

alexira [117]1 year ago

8 0

Answer:

Its too blurry, I can't see. Try re-posting the question with the image, but a little closer please.

Step-by-step explanation:

alexandr402 [8]1 year ago

6 0

Answer:

2 1/4

Step-by-step explanation:

3/4+2/4=5/4 or 1 1/4 + 1 = 2 1/4

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The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.The ma

Lana71 [14]

Answer:

The advertisement should use 16 minutes.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.

This means that $m = 3.5, \mu = \frac{1}{3.5} = 0.2857$

What number of minutes should the advertisement use?

The values of x for which:

$P(X > x) = 0.01$

$e^{-\mu x} = 0.01$

$e^{-0.2857x} = 0.01$

$\ln{e^{-0.2857x}} = \ln{0.01}$

$-0.2857x = \ln{0.01}$

$x = -\frac{\ln{0.01}}{0.2857}$

$x = 16.12$

Rounding to the nearest number, the advertisement should use 16 minutes.

5 0

2 years ago

(Sorry for it being so long, I just need help with these questions.)

sattari [20]

Answer:

1. c 2. a 3. b

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Help me ! i don't like mathhhh !

natali 33 [55]

4 (1/2 + 5/4) + -5

4 7/5 + -5

5 3/4 + -5

3/4

4 0

2 years ago

Read 2 more answers

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

2 years ago

How do you express your terms in pi

il63 [147K]

Answer:

To express your answer in terms of pi, simply refrain from substituting pi's numerical value for its symbol in the equation. That way, your answer will look like xπ where x is whatever number you come up with, and π is simply a placeholder.

4 0

2 years ago

Read 2 more answers