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Andreas93 [3]
3 years ago
8

9. If you randomly select one of the pieces of chocolate from the box, what is the probability that the piece will be filled wit

h fudge?
Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0

Answer:

Probability that the piece will be filled with fudge is 4/15 = 0.2667

Step-by-step explanation:

In the question, a box of chocolate contains 15 identical looking pieces. These chocolate pieces are filled in the following way: There are

3 caramel

2 cherry cream

2 coconut

4 chocolate whip

4 fudge

So this makes 3+2+2+4+4= 15

As the total number of chocolate pieces are 15

Let E be the event that the chocolate piece is filled with fudge

As we know Probability of an event is

P(E) = number of favorable outcomes / number of possible outcomes

Here the favorable is the number of pieces filled with fudge and number of possible outcomes are 15 as there are 15 types of chocolate pieces. So the probability that the piece selected will be filled with fudge is:

                        P(E) = 4/15 = 0.2667

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Your carry-on bag can weigh at most 40 pounds.Select an inequality that represents how much more weight x (in pounds) you can ad
andrey2020 [161]

Answer:

22 + x ≤ 40

x ≤ 18 pounds

Step-by-step explanation:

Maximum weight = 40 pounds

Initial weight = 22 pounds

Let Maximum weight that can be added = x

Mathematically, this means :

Initial weight + additional weight ≤ 40

That is ;

22 + x ≤ 40

Additional weight, x :

x ≤ 40 - 22

x ≤ 18 pounds

7 0
3 years ago
Data is collected to perform the following the hypothesis test:
nekit [7.7K]

Answer:

option f is right

Step-by-step explanation:

Given that data is collected to perform the following hypothesis test.

H_0: \mu = 5.5\\H_a: \mu >5.5

(right tailed test)

Sample mean = 5.4

p value = 0.1034

when p value = 0.1034 we normally accept null hypothesis.  i.e chances of null hypothesis true is  the probability of obtaining test results at least as extreme as the results actually observed during the test, assuming that the null hypothesis is correct

f) If the mean µ does not differ significantly from 5.5 (that is, if the null hypothesis is true), then the probability of obtaining a sample mean y as far or farther from 5.5 than 5.4 is .1034.

.

5 0
3 years ago
A bakery recorded the number of muffins and bagels it sold for a seven day period. For the data presented, what does the value o
atroni [7]

Answer:

Mean of bagel

Step-by-step explanation:

Working out each of the statically summaries Given in the option :

Mean of muffin :

Mean = Σx / n

n = sample size

(61 + 20 + 32 + 58 + 62 + 61 + 56) / 7

= 350 / 7

= 50

Mean of bagel :

Mean = Σx / n

n = sample size

(34 + 45 + 43 + 42 + 46 + 72 + 75) / 7

= 357 / 7

= 51

Hence, from the result obtained, we could see that the value of 51 summarizes the mean of bagel.

3 0
3 years ago
2. An online bookstore sells both print books and e-books (books in an electronic format). Customers can pay with either a gift
Fudgin [204]

Answer:

a.0.12

b. Dependent

Step-by-step explanation:

a. The two events mentioned here are " print book is purchases" and "customer using gift card". P(P)=0.6 and P(C)=0.2. we have to calculate the probability of print book selection for purchase and payment is made through gift card i.e. P(P∩C). P(P∩C)=P(P)*P(C)=0.6*0.2=0.12.

b. If P(A∩B)=P(A)*P(B), then the two events are independent. Here in the given situation two independent variables are "purchase of e-book" and  "payment made using gift card"

P(E)=0.4

P(G)=0.2

P(E and G)=P(E∩G)=0.1

P(E)*P(G)=0.4*0.2=0.08 that is not equal to 0.1. So, the two events "e-book is purchased" and "Payment made by customer using gift card" are not independent.

5 0
3 years ago
the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years​
Yuliya22 [10]

Answer:  125 g

<u>Step-by-step explanation:</u>

A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k

\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125

8 0
3 years ago
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