Jacob has two $1 bills, 2 dimes and 3 pennies in his pocket. How much money does jacob have in his pocket?

^ this week, The sparkling water was 4 for $5 , the regular was $1.69

S_A_V [24]

This week you paid $1.25 each because they were on sale for four for $5.

4 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

To make fruit punch for a party, we need 4 1⁄2 gallons of ginger ale, 1 gallon of strawberry juice, 2 3⁄4 gallons of frozen oran

lina2011 [118]

For this case we convert the mixed numbers to fractions:

$4 \frac {1} {2} = \frac {2 * 4 + 1} {2} = \frac {9} {2}\\2 \frac {3} {4} = \frac {4 * 2 + 3} {4} = \frac {11} {4}$

Now we add the fractions:

$\frac{9} {2}+1+ \frac {11} {4} + \frac {1} {8} =$

We find the least common multiple of the denominators. The l.c.m is 8, then we divide by each denominator and multiply the result by each numerator.

$\frac {36 + 8 + 22 + 1} {8} =\\\frac {67} {8} =$

8,375

Thus, the punch recipe implies 8.375 gallons. If we convert to a mixed number we have:

$8 \frac {3} {8}$ gallons

Answer:

$8 \frac {3} {8}$ gallons

4 0

3 years ago

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

3 years ago

In two or more complete sentences, identify the parent function, and describe the transformations that were applied to

pantera1 [17]

Answer:

Step-by-step explanation:

Transformed of the function has been given as,

f(x) = $\sqrt{(2x+6)}+3$

= $\sqrt{2}(\sqrt{x+3})+3$

Parent function of this transformed function = $\sqrt{x}$

Vertical stretch = By a factor of $\sqrt{2}$

Horizontal shift = 3 units to the left

Vertical shift = 3 units up

Now we can summarize these transformations as,

1). Parent function g(x) = $\sqrt{x}$ has been vertically stretched by a factor of $\sqrt{2}$

2). Followed by the translations of 3 units left and 3 units up.

3 0

3 years ago