2. José is using 12 brown tiles and

A package of hamburgers contains 8 patties and costs $7.50. Part A.) Luna has to buy at least 16 packages for an upcoming picnic

Rus_ich [418]

Answer:

Part A) $p\geq 128\ patties$

Part B) She will spend more than $\$142.50$

Step-by-step explanation:

Let

p-----> the number of hamburger patties

Part A) Luna has to buy at least 16 packages for an upcoming picnic

$p\geq 16*8$

$p\geq 128\ patties$

Part B) Suppose she actually needs more than 150 hamburgers. How much will she spend?

Let

c---------> the total cost

step 1

Divide 150 hamburgers by 8 (a package of hamburgers)

so

$\frac{150}{8} =18.75\ package$

round to the nearest whole number

$18.75=19\ package$ ----> the minimum number of packages

step 2

$c>19*\$7.50$

$c>\$142.50$

7 0

2 years ago

Callie was billed $416.48 for car repairs. She will pay the service center in 4 equal monthly installments from her bank account

Sav [38]

Answer:

her monthly payments will be $104.12

Step-by-step explanation:

she will pay off the entire bill after 4 months

416.48 - 104.12n = 0

-104.12n = -416.48

n = 4

6 0

1 year ago

Read 2 more answers

The singing club keeps sheet music in files. Each file holds 12 sheets. Of all the club's music, 2/3 fit into 3 files. How many

laila [671]

1 file -> 12 sheets

Now for 3 files -> 3*12 = 36 sheets, it takes up 2/3 of the total

So the total number of sheets: 36 / (2/3) = 54

6 0

2 years ago

Fiona wrote the linear equation y = x – 5. When Henry wrote his equation, they discovered that his equation had all the same sol

Oliga [24]

His equation could be written in quadratic form, which is ax^2+bx=c

7 0

1 year ago

Read 2 more answers

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

2 years ago