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earnstyle [38]
3 years ago
10

The ratio of students packing a lunch to those buying a lunch is 2:9. The number of students buying lunch is how many times the

number of students packing a lunch?
Mathematics
2 answers:
Gekata [30.6K]3 years ago
4 0

Answer:

4.5 9 divided by 2 = 4.5

Step-by-step explanation:

9 divided by 2 = 4.5

Marta_Voda [28]3 years ago
3 0
 4.5 times the number of students packing a lunch
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Round 411 to the nearest hundred​
Ghella [55]

Answer:

The answer is 400

5 0
2 years ago
A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two s
Rzqust [24]

Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black  & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:  

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

 P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

 P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

 P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black  one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

4 0
3 years ago
How do i solve this problem?
Eduardwww [97]

I think the first numerator cant be Z so I solved the question assuming it is 2;

Answer:

(x,y,z)=(1,5,7)

Hope this helps.

7 0
3 years ago
Is the number 15 prime or composite?
kramer
It is composite because it can be factored out as 5x3 or 15x1

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3 years ago
Help anyone plz?????
ryzh [129]

Answer:

.

Step-by-step explanation:

8 0
3 years ago
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