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maxonik [38]
2 years ago
6

Jane likes to exercise daily. The table shows the number of calories y she burns by exercising steadily for x minutes. How many

calories would she burn by exercising for 29 minutes?

Mathematics
2 answers:
Nataly_w [17]2 years ago
8 0
Do you have a picture of the table?
marusya05 [52]2 years ago
8 0

Answer:

<h2>319 cal/min.</h2>

Step-by-step explanation:

The table is attached.

You can observe in the table that there's no a value for 29 minutes.

That means we need to find the linear expression and then calculate the answer.

First, we need to find the ratio of change using two pairs of the given table and the following formula, (20,220) and (40,440) are the points

m=\frac{440-220}{40-20}=\frac{220}{20} =11

The ratio of change states that Jane burns 11 calories per minute. So, after 29 minutes, she will burn

11cal(29min)=319 cal/min

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Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the firs
strojnjashka [21]

Answer:

Answer has explained below.

Step-by-step explanation:

Consider the function is:

F(x) = 25x3 – 6x2 +7x -88

Differentiate with respect to x, we get

F’(x) = 25. 3x2 – 6.2x + 7

       = 75x2 – 12x +7

At x = 2, we have

F (2) = 25(2)3 – 6(2)2 + 7(2)-88

        =102

And f’(2) = 75(2)2 – 12 (2) +7

               =283

Now, calculate forward divided difference as:

xi + 1 = xi + h

        =2 + 0.25

        =2.25

F (xi + 1) = f (2.25) = 25 (2.25)3 – 6(2.25)2 +7(2.25) -88

                            =182.21

f’(2) = f(2.25) – f(2) / 0.25 = 182.21 – 102 / 0.25

                                             = 320.84

Єt = 283 – 320.8 / 283 = -13.36%

Now calculate backward divided difference:

Xi-1 = xi –h = 2 – 0.25 = 1.75

F(xi-1)= f(1.8) = 25 . (1.8)3 -6 (1.8)2 + 7 (1.8) – 88

                       = 50.96

F’(2) = f(2) – f(1.8) / 0.25 = 102 – 50.96 / 0.25 = 204.16

Єt = 283 – 204.16 / 283 = 27.86%

Finally, centered divided difference is obtain by inserting f(xi+1) and f (xi-1):

F’(2) = f(2.25) – f(1.8) /2 x 0.25 = 320.84 -50.96 / 0.5 = 539.68

Єt = 283 – 539.68 / 283 = -90.7%

6 0
2 years ago
The table shows different quantities of flour and the number of cookies that can be made with each quantity of flour. If the rel
nirvana33 [79]
3/4 = 6/x....3/4 = 6/8...notice that proportions are nothing but equivalent fractions

x = 8
4 0
2 years ago
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Marks age, X is six times his age 2 years ago. which equation represents the statement
FinnZ [79.3K]

x=6x-2 This would be your answer, hope this helps if you need any further explanation feel free to let me know and I would be happy to help you out.

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2 years ago
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During the early 20th century, in what way was the radio better than the phonograph in encouraging the concept of crosspollinati
ololo11 [35]

C. Phonograph were inferior in sound quality.

Step-by-step explanation:

Researches have shown that plants respond to the sounds of their pollinator.

Some plants begin to produce a greater amount of nectar and some begin to vibrate to attract the attention of the pollinators towards them.

Greater the intensity of the sound of the pollinator, the faster the reaction will be.

Phonographs have a sound of lower intensity when compared to the intensity of sound produced by the radio.

Thus the radio was better than the phonograph in encouraging the concept of cross-pollination.

6 0
2 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
2 years ago
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