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Arte-miy333 [17]
3 years ago
5

will ate a snack with 68.44 total calories.If the chip he ate were 39.94 calories, how many calories were in the rest of his sna

cks
Mathematics
1 answer:
jasenka [17]3 years ago
8 0
Subtract the total with what you have left in the chip bag. 68.44-39.94= 28.94
You might be interested in
Write the first five terms of the sequence in which the nth term is
ziro4ka [17]

Answer:

C) 2,6,24,120,720

Step-by-step explanation:

Here the n-th term An = \frac{(n +2)!}{n + 2}

A1 is the first term

A1 = (1 + 2)!/(1 + 2)

= 3!/3

A1 = 2

A2 is the second term

A2 = (2 +2)!/(2 +2)

= 4!/4

A2 = (1*2*3*4) /4

A2 = 6

A3 is the third term

A3 = (3 + 2)!/(3 +2)

A3 = 5!/5

A3 = 24

A4 is the fourth term

A4 = (4 + 2)!/(4 + 2)

A4 = 6!/6

A4 = 120

A5 is the fifth term

A5 = (5 + 2)!/(5 +2)

A5 = 7!/7

A5 = 720

Answer: C) 2,6,24,120,720

Thank you.

8 0
3 years ago
If D is the midpoint of CE, CD= 9x-7, and DE= 3x+17, find CE
tatuchka [14]
Hello!

Since D is the midpoint and the two equations are and both sides of point D the equations equal each other

9x - 7 = 3x + 17

Now you solve it algebraically

Add 7 to both sides

9x = 3x + 24

Subtract 3x from both sides

6x = 24

Divide both sides by 6

x = 4

Now we put this into both equations and add them

9(4) - 7 = 29
3(4) + 17 = 29

29 + 29 = 58

The answer is 58 units

Hope this helps!
3 0
3 years ago
Can someone please help me with this question? I have been stuck on it for a while now​
frez [133]

Answer:

<em>w = 12 cm  </em>and <em> l = 20 cm</em>

Step-by-step explanation:

3 0
3 years ago
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
Hilda has $210 worth of $10 and $12 stock shares. The number of $10 shares is five more than twice the number of $12 shares. How
klasskru [66]
Let's first define variables.
 y: number of shares of $ 10
 x: number of shares of $ 12
 We write the system of equations:
 10x + 12y = 210
 y = 2x + 5
 Solving the system:
 x = 75/17
 y = 235/17
 Answer: 
 she has: 
 75/17 shares of $ 12 
 235/17 shares of $ 10

6 0
3 years ago
Read 2 more answers
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