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3 years ago

PLEASE HELP ME ASAP, and pleaseeee dont put a link or I will report you asap. screenshot of the assignment is attached

Mathematics

1 answer:

blagie [28]3 years ago

7 0

Answer:

Step-by-step explanation:

(1 + 6) × 12 ÷ 2 = 42

area of a trapezoid:

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Y+15 <3 for inequality

Hitman42 [59]

Hello there!

y + 15 < 3

Start by subtracting 15 on both sides

y + 15 - 15 < 3 - 15

y < -12

As always, it is my pleasure to help you guys on here. Let me know if you have any questions.

7 0

3 years ago

a student has a 73% of being let out of class late. if he leaves class late. if he leaves class late, there is a 23% chance that

Elza [17]

Answer:

16.79 %

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1

muminat

Answer:

$Probability = \frac{1}{24}$

Step-by-step explanation:

Given

$Oranges = 3$

$Apples = 5$

$Banana = 2$

$Total = 10$

Required

Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

$Probability = P(Orange) * P(Apple) * P(Banana)$

$Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}$

<em>The difference in the numerator is as a result of picking the fruit without replacement</em>

$Probability = \frac{30}{720}$

$Probability = \frac{1}{24}$

6 0

2 years ago

What is 1 and 5/6 + 4 and 3/6

Simora [160]

The answer is 6 1/3 because 1+4= 5 and 3/6 + 5/6 is 8/6 which simplify a to 1 and 1/3 and 5+ 1 1/3= 6 1/3

3 0

3 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

2 years ago