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3 years ago

How many different menus showing 10 main courses can a restaurant make if it has 15 main courses from which to choose?

1 answer:

6 0

The number of ways or different menus containing 10 main courses that the restaurant can make out of the 15 main courses is calculated through the combination.

15C10

This is read as "combination of 15 taken 10" which can be calculated through the equation,

nCr = n! / ((r!) x(n - r)!)

In this item, n is equal to 15 and r is equal to 10.

Substituting the known values to the equation above,

15C10 = 15! / (10!)x(10 - 5)! = 3003

Thus, there are 3003 different menus that the restaurant can choose from.

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3 years ago

Evaluate -a - 12a^2 for a = -1 a. -13 b. 13 c.-11

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Let x be a positive integer. If (-3)ºx (-3)* = (-3)14, what is x?

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Answer:

$-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14 : x = \frac{840}{180^{\circ \:}}$

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$x = -267.38030...$

Step-by-step explanation:

$-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14$

Remove Parentheses: $(-a) = -a$

$-3^{\circ \:}x\left(-3\right)=-3\cdot \:14$

Multiply the numbers: $3 * 14 = 42$

$-3^{\circ \:}x\left(-3\right)=-42$

Divide both sides by: $-3^{\circ \:}\left(-3\right)$

$\frac{-3^{\circ \:}x\left(-3\right)}{-3^{\circ \:}\left(-3\right)}=\frac{-42}{-3^{\circ \:}\left(-3\right)}$

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$x=-\frac{840}{180^{\circ \:}}$

Hope I helped. If so, may I get brainliest and a thanks?

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3 years ago

Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp

aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

$F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}$

Using a statistic software I've calculated the test

$F_{H_0}= 1.41$

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

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

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03$

$t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15$

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0

3 years ago