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3 years ago

7

What is the probability of rolling a 4 (four) on the 1st die.... and a 1 (one) on the 2nd die. Yes.. it matters which die is whi

ch !?!?!?!? Help pleaseee

1 answer:

disa [49]3 years ago

4 0

If you're doing probabilities, there is a 1:6 chance on both (im guessing dont trust me i did probabilities a long time ago)

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The temperature at 6:00 pm was 24°. At midnight the temperature 36° lower than 6:00pm . What was the temp at midnight?

horrorfan [7]

Answer:

-12°

Step-by-step explanation:

Man... I just did 36 minus 24, and since it says 36° lower I decided it was -12 degrees (24 - 36 is more accurate but I find it easier just to do it the other way)

4 0

3 years ago

Write this number in expanded form 382,706

77julia77 [94]

Answer:

Expanded form : 300,000 +80,000 +2000 +700 +6.

Step-by-step explanation:

Given : 382,706.

To find : Write this number in expanded form.

Solution : We have given 382,706.

We can see 3 is at hundred thousand place = 300,000

8 is at ten thousand place = 80,000.

2 is at thousand place = 2000.

7 is at hundred place = 700.

6 is at ones place = 6

Then

Expanded form : 300,000 +80,000 +2000 +700 +6.

Therefore, Expanded form : 300,000 +80,000 +2000 +700 +6.

5 0

3 years ago

Read 2 more answers

In a group of Explore students, 38 enjoy video games, 12 enjoy going to the movies and 24 enjoy solving mathematical problems. O

Elodia [21]

Answer:

The number of students that like only two of the activities are 34

Step-by-step explanation:

Number of students that enjoy video games, A = 38

Number of students that enjoy going to the movies, B = 12

Number of students that enjoy solving mathematical problems, C = 24

A∩B∩C = 8

Here we have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) -n(A∩C) + n(A∩B∩C)

= 38 + 12 + 24 - n(A∩B) - n(B∩C) -n(A∩C) + 8

Also the number of student that like only one activity is found from the following equation;

n(A) - n(A∩B) - n(A∩C) + n(A∩B∩C) + n(B) - n(A∩B) - n(B∩C) + n(A∩B∩C) + n(C) - n(C∩B) - n(A∩C) + n(A∩B∩C) = 30

n(A) + n(B) + n(C) - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 3·n(A∩B∩C) = 30

38 + 12 + 24 - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 24 = 30

- 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) = -68

n(A∩B) + n(B∩C) + n(A∩C) = 34

Therefore, the number of students that like only two of the activities = 34.

8 0

3 years ago

3/7 to 6/7. what is the percent of change

stepan [7]

Percent change = (new - old)/old * 100
(6/7 - 3/7)/(3/7) * 100
3/7 * 7/3 * 100
1 * 100
100%

4 0

3 years ago

Read 2 more answers

The Venn diagram shows the results of two events resulting from rolling a number cube.

gulaghasi [49]

Answer: $P(A\cap B)=\frac{1}{3}$

$P(A)=\frac{1}{3}$

$P(B)=1$

$P(A|B)=\frac{1}{3}$

Step-by-step explanation:

From the given figure it can be seen that

Total number on cube= n(S)=6

Intersection of A and B = n(A ∩ B)= 2

Therefore,

$P(A\cap B)=\frac{n(A\cap B)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Also, The number of elements in A = 2

Therefore,

$P(A)=\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}$

Similarly, The number of elements in B= 6

Therefore,

$P(B)=\frac{n(B)}{n(s)}=\frac{6}{6}=1$

The formula to find the conditional probability is given by :-

$P(A|B)=\frac{P(A\cap B)}{P(B)}\\\\\Rightarrow P(A|B)=\frac{\frac{1}{3}}{1}=\frac{1}{3}$

4 0

3 years ago

Read 2 more answers