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3 years ago

Can someone help me with this problem ASAP

Mathematics

1 answer:

o-na [289]3 years ago

7 0

28÷7=4 and 35÷7 is 5 so they can both be divided by seven so 7 students should be in each group

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Solve For U<br> -4 +2u = 6

Drupady [299]

Answer:

u=5

Step-by-step explanation:

-4 +2u = 6

2u=10

u=5

6 0

4 years ago

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Please help me solve 3 (1+2v)-3(1+4v)

Vilka [71]

Answer:

18v

Step-by-step explanation: See the attached file! Enjoy. If you need explanation, just tell me! Id be glad to answer more than answer per question!

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4 years ago

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A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10

kompoz [17]

Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$

6 0

3 years ago

What the answer to 1-5+1*(4*4-31)*8

DiKsa [7]

Answer:

-124

Step-by-step explanation:

1-5+(-15)8

1-5-120

-4-120

-124

4 0

3 years ago

2.9y=11.6 what does y equal?

zhuklara [117]

Answer:

Step-by-step explanation:

6 0

4 years ago