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3 years ago

6

A local pizza restaurant offers a gold card for loyal customers. The gold card costs $25 for a year. With a gold card, each larg

e pizza is a discounted cost of $12. If customers don’t have a gold card, a large pizza costs $18. How many pizzas must you purchase in a year for the gold card to be cheaper?

1 answer:

VMariaS [17]3 years ago

8 0

Answer:

5 pizzas

Step-by-step explanation:

18-12=6 each pizza with card saves 6$

$25 card cost÷$6 savings per pizza=4.16 pizza to break even. round up since cant buy fraction of pizza.

check

5 pizza with card 12 *5=60

60 + 25 = 85 spent in total

$5 *$18 = $90, no discount

4 pizza would be 73 with discount and 72 without

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Answer:

Total space in the box is 75.2 cm³.

Step-by-step explanation:

Cone:

It is three dimension shape.
It has a vertex.
The slant height of a cone $l^2=r^2+h^2$ , $l$ = slant height, r= radius of the base, h= height
The lateral surface area is πrl
The volume of a cone is $\frac13 \pi r^2h$ .

The artificial carrot that are cone shaped with a radius 1 cm and a height of 6 cm.

So, the volume of the cone is = $\frac13 \pi r^2h$

$=\frac13 \pi (1)^2.6$

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The volume 1 dozen carrots is = (12×2π) cm³

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The volume of the box= The volume of the 1 dozen carrot.

Total space in the box is 75.2 cm³.

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3 years ago

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

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vovikov84 [41]

Note :- If sample size n > 30 OR Population standard deviation σ is given then We use Z test.

If sample size n < 30 AND Population standard deviation σ is unknown then we use t test.

In this question we are not given Population standard deviations σ1 and σ2.

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Therefore we use t test.

Given :- Assume the population standard deviations are equal. ( σ1 = σ2)

then Degrees of freedom = n1 + n2 - 2 = 14 + 13 - 2 = 25.

(kindly find the image attached with this solution)

Answer :- Therefore 25 degrees of freedom are used to find the t critical value.

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damaskus [11]

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Answer:

see attached

Step-by-step explanation:

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