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Ivahew [28]
3 years ago
13

1) El ritmo cardíaco de un gato ( en latidos por minuto) es función lineal de su temperatura corporal (en grados Celsius). En co

ndiciones normales un gato con 37°C tiene un ritmo cardíaco de 200 pulsaciones por minuto y de 150 si su temperatura es de 32°C.
Mathematics
1 answer:
olga55 [171]3 years ago
6 0

Answer:

y=10x-170

Step-by-step explanation:

This problem can be modeled by a linear function, where we have two points (37°C, 200) and (32°C, 150).

Notice that the temperature is the independent variable, because the heartbeat of the cat won't change the temperature, it works in the opposite way.

First, let's find the constant ratio of change.

m=\frac{150-200}{32-37}=\frac{-50}{-5}=10

This means the hearbeat of the car changes 10 pulses per minute each 1 celsius degree.

Now, we use the point slope formula to find the equation

y-y_{1} =m(x-x_{1} )\\y-150=10(x-32)\\y=10x-320+150\\y=10x-170

Theerefore, the function that represents this situation is

y=10x-170

Where x represents temperature and y represents hearbeats.

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Airplane descended 375 m in 10 seconds. What was the airplane’s rate of change in elevation (how many meters does the airplane d
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37.5 m/s

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Find the range for the given domain: {-2, -1, and 0} and the function is y= 3x^2
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  {0, 3, 12}

Step-by-step explanation:

Put the domain values in the function and evaluate. The result is the range values.

  y = 3{-2, -1, 0}^2 = 3{4, 1, 0} = {12, 3, 0}

For the given domain, the range is {0, 3, 12}.

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<em>Additional comment</em>

Neither the domain nor range values need to be put in any particular order. However, it is often convenient for them to be arranged from least to greatest.

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3 years ago
The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So
tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 \times 10^{-6}.

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........

where, n = number of trials (samples) taken = 18 Southwest flights

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Let X = <u><em>Number of flights that arrive late</em></u>.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X \geq 13)

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= \binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}

= \binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}

= 2.5196 \times 10^{-6}.

7 0
3 years ago
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