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Sindrei [870]
2 years ago
7

A pet store donated 50 pounds of food for adult dogs, puppies, and cats to an animal shelter. 19 and 3/4 pounds was adult dog fo

od and 18 and 7/8 pounds was puppy food. How many pounds of cat food did the pet store donate?
Mathematics
2 answers:
Sati [7]2 years ago
8 0
So first we will add together the fractions. 3/4 is equal to 6/8, so we will add 6/8 to 7/8 and get 13/8, or 1 5/8.  Then we can add together the whole numbers, 19 and 18, to get 37. Then add 37 + 1 5/8 and get 38 5/8. Lastly, we subtract that amount from 50 to get a finale number of 11 3/8. The pet store donated 11 3/8 pounds of cat food! Hope this helps!
madam [21]2 years ago
3 0
50 - (19 3/4 + 18 7/8) = the cat food
50 - (19 6/8 + 18 7/8) 
50 - (37 13/8) =
50 - (38 5/8) = 
11 3/8 lbs of cat food <==
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The national electrical code allows a maximum voltage drop of 3% on branch circuits. What is the maximum allowable drop on a 240
Andre45 [30]

Answer:

  7.2 volts

Step-by-step explanation:

3% of 240 is ...

  0.03 × 240 = 3 × 2.40 = 7.20

The maximum allowable drop on a 240-volt circuit is 7.2 volts.

7 0
2 years ago
If a conditional statement is “If the sun shines, people will be in good moods”, what is the statement “If the sun does not shin
77julia77 [94]

Answer:

C. Inverse

Step-by-step explanation:

The second statement is the inverse, because in an inverse, the hypothesis and the conclusion are both negated. In this case, the sun doesnt shine and the people wont be in a good mood, as opposed to the sun will shine, and the people will  be in a good mood.

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1 year ago
A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa
sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_A- \mu_B = 0

Alternative hypothesis: \mu_A -\mu_B \neq 0

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a two tailed test the p value is given by:

p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0
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Kipish [7]

Answer:

Step-by-step explanation:

model C

7 0
2 years ago
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Goshia [24]
First we need to identify if the data is qualitative or quantitative.

The data is average number of people living in the homes.

Qualitative data as its name indicates is an attribute or characteristic. It can not be measured e.g color. Quantitative data is such a data which can be counted or measured.

Since the average number of people can be counted and measured, the data is Quantitative.

In an observational study the individuals are observed. In the given case, Kira did not observed the individuals to gather the data, rather she used an Online resource to gather the data.

Therefore, the correct answer will be:
Kira used published data that is quantitative.
4 0
2 years ago
Read 2 more answers
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