All categories

3 years ago

Why did the troll give us the clues

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

lol

Step-by-step explanation:

im laughing in class

You might be interested in

Daniel bought four bags of potatoes. The weight of the first bag was 2.6 pounds. The second bag weighed 0.4 pound less than twic

ipn [44]

Answer:

Average weight of the bags of potatoes is 3.8 pounds.

Step-by-step explanation:

Given:

Weight of the first bag, $W_{1}=2.6$ pounds.

Weight of the second bag is 0.4 pounds less than twice the weight of first bag. This means,

$W_{2}=2W_{1}-0.4=2(2.6)-0.4=5.2-0.4=4.8\textrm{ pounds}$

Weight of the third bag is 0.6 pounds more than that of the first bag. This means,

$W_{3}=W_{1}+0.6=2.6+0.6=3.2\textrm{ pounds}$

Weight of the fourth bag is 0.3 pounds less than that of the second bag. This means,

$W_{4}=W_{2}-0.3=4.8-0.3=4.5\textrm{ pounds}$

Therefore, the average weight of the bags of potatoes is given as the sum of all the weights and then divide the sum by 4. Therefore,

$W_{average}=\frac{W_{1}+W_{2}+W_{3}+W_{4}}{4}=\frac{2.6+4.8+3.2+4.5}{2}=\frac{15.1}{4}=3.775\approx 3.8$

Therefore, average weight of the bags of potatoes is 3.8 pounds.

8 0

3 years ago

BRANILIEST OFFER!!what is the median of this data set?<br> 25, 8, 10, 35,45, 5, 40, 30. 20.

ivann1987 [24]

Median is the middle of the number set so
5 8 10 20 25 30 35 40 45
its 25

4 0

3 years ago

Read 2 more answers

Prove the divisibility of the following numbers:

yan [13]

$8^{10}-8^9-8^8=8^8(8^2-8-1)=8^8(55)$

$81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}(3^2-3-1)=3^{26}(5)=3^{24}(45)$

8 0

3 years ago

Select the correct answer. What is the solution to the equation? 3(x + 9)^3/4 = 24

bazaltina [42]

Answer:

C~7

Step-by-step explanation:

7 0

3 years ago

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

3 years ago