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valkas [14]
2 years ago
13

The energy released by metabolism of 1 average candy bar is 1 x 10^6 joules.

Mathematics
1 answer:
Elis [28]2 years ago
3 0

Answer: 32

Step-by-step explanation:

Given: The energy needed for 1 hour of running for an adult =4.5\times10^6\ joules.

Therefore. the energy needed for seven hours of running==7\times4.5\times10^6\ joules.=31.5\times10^6\ joules

Since, he energy released by metabolism of 1 average candy bar =1\times10^6\ joules.

Therefore, the number of candy bars an adult need to eat  to supply the energy needed for seven hours of running.=\frac{31.5\times10^6}{1\times10^6}=31.5\approx32

Hence, the adult need to eat 32 candies to supply the energy needed for seven hours of running.

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Margie's car can go 3232 miles on a gallon of gas, and gas currently costs $4$4 per gallon. How many miles can Margie drive on $
vaieri [72.5K]
She should be able to drive 160 miles
4 0
3 years ago
Find the sum of a finite geometric sequence from n = 1 to n = 8, using the expression −2(3)^n − 1.
Kazeer [188]
The sum if the geometric sequence given by:
an=-2(3)^(n-1)
will be:
when:
n=1
an=-2
when n=2
a2=-6
when n=3
a3=-18

when n=4
a4=-54

when n=5
a5=-162

when n=6
a6=-486

when n=7
a7=-1458

when n=8
a8=-4374

thus the summation of the term will be:
Sn=(-4374+-1458+-486+-162+-54+-18+-6+-2)
Sn=-6560
the answer is -6560

6 0
3 years ago
Genevieve wants to verify that }(5x-20)-(4x-8) is equivalent to x. Which procedure can Genevieve follow to determine if
Gnom [1K]

They aren't equivalent

<h2>Explanation:</h2>

In order to determine whether the expression:

(5x-20)-(4x-8)

is equivalent to:

x

Genevieve needs to follow these steps:

Step 1. Get rid parentheses

5x-20-4x+8

Step 2. Combine like terms

5x-4x-20+8

Step 3. Solve

x-12

So:

x\neq x-12

Therefore, these two expressions aren't equivalent.

<h2>Learn more:</h2>

Writing expressions: brainly.com/question/13894833#

#LearnWithBrainly

4 0
3 years ago
Read 2 more answers
A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
KiRa [710]

Answer:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

4 0
3 years ago
‼️help please thank you ☺️
Kay [80]
The answer is true...
4 0
3 years ago
Read 2 more answers
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