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2 years ago

Pls help. No files pls. Will give brainliest.

Mathematics

1 answer:

nordsb [41]2 years ago

6 0

To find f(-20), first figure out which piece x = -20 fits with.

Since -20 < -12, x = -20 first in the domain used by the third piece.

For f(-20), treat this function as if it was just f(x) = 3x-7.

f(-20) = 3(-20) -7

= -60 - 7

= -67

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Solve the inequality -12≤2x-4<10

tia_tia [17]

I hope this answers your question !

4 0

2 years ago

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

3 years ago

In an international film festival, a penal of 11 judges is formed to judge the best film. At last two films FA and FB were consi

valentina_108 [34]

Answer:

The answer is "0.4329 ".

Step-by-step explanation:

P( three in favor of FA)

Select 3 out of 6 FA supporters then select 2 out of 5 FB supportive judges

$=\frac{^{6}_{C_{3}}\times ^{5}_{C_{2}}}{^{11}_{C_{5}}}\\\\=\frac{\frac{6!}{3!(6-3)!}\times \frac{5!}{2!(5-2)!}}{\frac{11!}{5!(11-5)!}}\\\\=\frac{\frac{6!}{3! \times 3!}\times \frac{5!}{2! \times 3!}}{\frac{11!}{5! \times 6!}}\\\\=\frac{\frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1\times 3!}\times \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}}{\frac{11 \times10 \times 9 \times 8 \times 7 \times 6! }{5 \times 4 \times 3 \times 2 \times 1 \times 6!}}\\\\$

$=\frac{ (5 \times 4) \times(5 \times 2)}{(11 \times 3 \times 2 \times 7 )}\\\\=\frac{ 20 \times 10 }{(11 \times 42)}\\\\=\frac{ 200 }{462}\\\\=\frac{100 }{231}\\\\=0.4329$

6 0

2 years ago

12598 round to the nearest thousand

MakcuM [25]

12598 rounded to the nearest thousands is 1260

5 0

3 years ago

Read 2 more answers

A survey was given to 339 people asking whether people like dogs and/or cats.

lina2011 [118]

Answer:

Step-by-step explanation:

Think of it as a Venn diagram. One circle is the people who like dogs, and one circle is the people who like cats. The overlap is people who like both dogs and cats.

190 people in the survey said they like dogs. That includes the people who like both dogs and cats.

141 people in the survey said they like cats. That includes the people who like both dogs and cats.

If we simply add the two numbers together, we'll be counting the overlap twice. So to find the total number of people who like dogs or cats, we have to subtract one overlap.

dogs or cats = 190 + 141 − x

Therefore:

190 + 141 − x + 88 = 339

419 − x = 339

x = 80

80 people said they liked both cats and dogs.

5 0

2 years ago