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Hoochie [10]
4 years ago
9

A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccol

i croissants. How many ways are there to choose
a)a dozen croissants?

b)three dozen croissants?

c)two dozen croissants with at least two of each kind?

d)two dozen croissants with no more than two broccoli croissants?

e)two dozen croissants with at least five chocolate croissants and at least three almond croissants?

f)two dozen croissants with at least one plain croissant, at least two cherry croissants, at least three chocolate croissants, at least one almond croissant, at least two apple croissants, and no more than three broccoli croissants?
Mathematics
1 answer:
slamgirl [31]4 years ago
5 0

Answer:

A. 6 188

B. 749 398

C. 6 188

D. 52 975

E. 20 349

F. 11 316

Explanation:

(a) The shop has 6 types of croissants of which a dozen(12) has to be selected

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(b) The shop has 6 types of croissants of which three dozen(36) has to be selected

Therefore n=6, r=36

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+36-1, 12) = C(41, 36) = 41!÷ 36!(41-36)! = 41!÷36! 5! = 749 398

(c) The shop has 6 types of croissants of which two dozen(24) has to be selected

Let us first select 2 of each kind which 12 croissants in total. Then we still need to select the remaining 12 croissants

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(d) The shop has 5 types of croissants of which two dozen(24) has to be selected

Therefore n=5, r=24

Repetition of croissants is permitted

And C(n+r-1, r)

C(5+24-1, 24) = C(28, 24) = 28!÷ 24!(28-24)! = 28!÷24! 4! = 20 475

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Suppose two bicyclists start at the same location.
DerKrebs [107]

Answer:

Step-by-step explanation:

See attachment.

If both cyclists travel for the same time and speed, they will have travelled the same distance.  Since one is headed north and the other east, we can see that the distance between them in one hour is the hypotenuse of a right triangle.  Each leg has distance x.  We can say x^2 + x^2 = (3\sqrt{2})^2

2x^2 =1 8

x^2 = 9

x = 3

They both rode 3 miles.

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3 years ago
PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!
Paraphin [41]

the overhead, fixed costs, is $1000.

the variable cost is $200, so if she sells "x" tiles, then her costs is 200*x or 200x.

total costs then will just be C(x) = 200x + 1000.

namely both costs added together.

Revenue is just how much it's been taken in from the sales, since each tile is selling for $240, then the revenue is simply 240*x or just R(x) = 240x.

break even point is when those two guys equal each other

\bf \stackrel{R(x)}{240x}=\stackrel{C(x)}{200x+1000}\implies 40x=1000\implies x=\cfrac{1000}{40}\implies x=25

what if the cost for each tile were $220, 20 bucks more?

\bf \stackrel{R(x)}{240x}=\stackrel{C(x)}{220x+1000}\implies 20x=1000\implies x=\cfrac{1000}{20}\implies x=50

well, if the cost for each tile is $200, her break-even point is at 25 units, namely once she has sold 25 tiles, she's has lost nothing, has won nothing either, but hasn't lost anything.

if the cost is however $220 instead, her break-even point comes much later, at 50 units, so she'll have to sell more tiles in order to not have any losses, so she's worse off if the cost is more or course.

4 0
3 years ago
Consider the following data set. The data were actually collected in pairs, and each row represents a pair.
Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

a.

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}

Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51

Sa= 2.12

t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875

The p-value for the two-tailed test is:

P(t_{18}≤-0.875) + P(t_{18}≥0.875)= P(t_{18}≤-0.875) + (1 - P(t_{18}≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

b.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}

t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05

The p-value for the two-tailed test is:

P(t_{9}≤-2.05) + P(t_{9}≥2.05)= P(t_{9}≤-2.05) + (1 - P(t_{9}≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

c.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!

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3 years ago
Luann predicted that if a certain number of vehicles passed her by,42 of them would be SUVs.What was that certain number of vehi
sweet-ann [11.9K]

Answer:

Are there multiple choice answers? because id guess 50. if its wrong im sorry...

Step-by-step explanation:

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3 years ago
Pilar says that the expression 3/4 divided by 1/2 has a quotient greater than 1, and the expression 1/2 divided by 3/4 has a quo
kiruha [24]
The statement is correct.

3/4 divided by 1/2 can be written as 3/4 x 2/1, because even dividing fractions you can flip the second one and multiply instead.

This gives 6/4, which can be written as 3/2. This is top heavy, meaning it is greater than 1. It could also be said as 3 divided by 2, and half of 3 is 1.5, greater than 1.

For the second one, 1/2 divided by 3/4 can be written as 1/2 x 4/3, which is 4/6, also written as 2/3.

2/3 is less than 3/3, making it less than 1.

This means Pilar is correct for both cases as the former is indeed greater than 1 and the latter is less than.
3 0
3 years ago
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