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ASHA 777 [7]
4 years ago
15

At a carnival, you spin a spinner that has four equal-sized sections, each a different color: green, yellow, red, and blue. If y

ou land on green, you win 2 points. If you land on yellow, red, or blue, you lose 1 point. Write the expected value equation. E (V) = StartFraction 1 Over a EndFraction (2) + StartFraction 1 Over b EndFraction (c) + StartFraction 1 OVer d EndFraction (e) + StartFraction 1 Over f EndFraction (g) a = b = c = d = e = f = g =

Mathematics
2 answers:
Anon25 [30]4 years ago
6 0

Answer:

A 4

B 4

C -1

D 4

E -1

F 4

G -1

-0.25 points

lose 250 points

7nadin3 [17]4 years ago
3 0

Answer:

a = 4

b = 4

c = -1

d = 4

e = -1

f = 4

g = -1

Step-by-step explanation:

To get the expected value equation you have to multiplicate the probability of each outcome by its reward and add them. The probability to get each color is 1/4 because the spinner has four equal-sized sections. If you win money, your reward is positive, but if you lost it, your reward is negative. Then:

expected value equation. E (V) = (1/4)*2 + (1/4)*(-1) + (1/4)*(-1) + (1/4)*(-1) = -0.25

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Kevin can carry a basket 5 feet Rachel can carry it three feet farther than Kevin Daniel can carry the basket half as far as Rac
OlgaM077 [116]
Yes. Kevin can carry the basket 5 ft. Rachel can carry it 3 feet more than Kevin (5+3=8). Daniel can carry the basket half the length of Rachel (8/2=4). The total distance they can carry the basket is 5+8+4=17. 17 is greater than 15, so yes they can carry it 15 feet.
8 0
3 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

5 0
3 years ago
Help ASAP i need help
Reptile [31]

Answer:

Isosceles Right Triangle

Step-by-step explanation:

We look at the sides first. Two sides are equal, which means that the triangle is isosceles (equilateral means three equal sides, and scalene means no sides equal). Then we look at the only angle indicated, which is the right angle. Thus, we come to the conclusion that the triangle is an Isosceles Right Triangle.

3 0
3 years ago
Help please?!
Furkat [3]
Well from what i know here's what i would put........
-4 * -2 = 8 - 2 = 6
-4 * -1 = 4 - 2 = 2
-4 * 0 = 0 - 2 = -2
so it does represent a function......I really hope that helped!!

8 0
3 years ago
A yogurt company claims that it prints a free yogurt coupon under a randomly selected 20% of its lids. A loyal customer purchase
Alona [7]

Answer:

The customer can conclude that the company's claim is correct

Step-by-step explanation:

The percentage of lids that has a free yogurt coupon = 20%

The number of cups a loyal customer purchases = 85 yogurt cups

The number of cups that contained a coupon = 12 (14.1%)

The confidence interval performed = 99% confidence interval for the proportion of yogurt cups containing coupon codes

The interval obtained = (0.044, 0.238)

Therefore, the range of proportion within which the true proportion exists is 0.044 < \hat p < 0.238

The range of percentage within which the true percentage exist is therefore;

0.044 × 100 = 4.4%  < \hat p × 100 < 0.238 × 100 = 23.8%

Given that the possible true percentage of lids that has a coupon is between 4.4% and 23.8% at 99% confidence level, the customer can conclude that only 12 of his yogurt cup contained coupon by chance and that the company's claim is correct.

4 0
3 years ago
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