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3 years ago

15

There are two spinners containing only red and orange slices. Spinner A has 7 red slices and 13 orange slices. All the slices ar

e the same size. Spinner B has 6 red slices and 6 orange slices. All the slices are the same size. Each spinner is spun. List these events from least likely to most likely. Event 1 : Spinner B lands on a red slice. Event 2 : Spinner A lands on a red or orange slice. Event 3 : Spinner A lands on a red slice. Event 4 : Spinner B lands on a black slice.

1 answer:

yan [13]3 years ago

8 0

Event 4 (impossible), Event 3 (probability is 7/20), Event 1 (probability is 6/12 which is greater than 7/20), and Event 2 (certain to happen)

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In The autumn 40% of the leaves on a tree are yellow and 24% of them are green the remaining 72 leaves are half yellow how many

Charra [1.4K]

Percentage of yellow leaves on a tree during Autumn = 40%

Percentage of green leaves on a tree during Autumn = 24%

Percentage of half yellow leaves on the tree during Autumn :

$= \tt100 - 40 + 24$

$= \tt 100 - 64$

$= \tt 36 \%$ Thus, the percentage of half yellow leaves on the tree during Autumn = 36%

Number of half yellow leaves on the tree = 36%

Let the total number of leaves on the tree be x.

Which means :

$= \tt36 \% \: of \: x = 72$

$= \tt \frac{36}{100} \times x = 72$

$= \tt \frac{36 \times x}{100} = 72$

$= \tt \frac{36x}{100} = 72$

$= \tt 36x = 72 \times 100$

$= \tt36x = 7200$

$= \tt x = \frac{7200}{36}$

$\color{plum} = \tt \: x = 200$

Thus, total number of leaves on the tree = 200

Number of yellow leaves on the tree :

$= \tt40 \% \: \: of \: \: 200$

$= \tt \frac{40}{100} \times 200$

$= \tt \frac{40 \times 200}{100}$

$= \tt \frac{8000}{100}$

$\color{plum} \tt = 80 \: yellow \: \: leaves$

Thus, total number of yellow leaves on the tree = 80

Number of green trees on the tree :

$= \tt24 \% \: \: of \: \: 200$

$= \tt \frac{24}{100} \times 200$

$= \tt \frac{24 \times 200}{100}$

$= \tt \frac{4800}{100}$

$\color{plum} = \tt 48 \: green \: \: leaves$

Thus, the total number of green leaves on the tree = 48

Since the sum of all types of leaves form 200[80+48+72=200], we can conclude that we have found out the correct number of each type of leaf.

▪︎Therefore, <em>the total number of leaves on the tree = 200</em>

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3 years ago

Three people are running for president of a class. The results of a poll indicate that the first candidate has an estimated 37%

lianna [129]

Answer:

0.19

Step-by-step explanation:

The are three candidate running for president and we know that probability of winning for first candidate and the probability of winning for second candidate and we have to find the probability of winning for third candidate

P(C1)=0.37

P(C2)=0.44

P(C3)=?

We know that sum of probabilities is always 1. So,

P(C1)+P(C2)+P(C3)=1

0.37+0.44+P(C3)=1

P(C3)=1-0.37-0.44

P(C3)=0.19

Thus, the probability of winning for third candidate is 0.19.

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3 years ago

In which equation is the constant of proportionality 5?

Finger [1]

Answer:b.y=5x
this is the answer to your question

8 0

3 years ago

Suppose that theta is an angle in standard position whose terminal side Intersects the unit circle at (-11/61, -60/61)

Blizzard [7]

Answer:

The exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

Step-by-step explanation:

The components of the unit vector are $x = -\frac{11}{61}$ and $y = -\frac{60}{61}$ . Since $r = 1$ , then $x = \cos \theta$ and $y = \sin \theta$ . By Trigonometry, tangent and secant can be calculated by the following expressions:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{x}$

Now, the exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

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NARA [144]

The first one is the answer

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