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Lubov Fominskaja [6]
3 years ago
13

What's the answer to the problem

Mathematics
1 answer:
zvonat [6]3 years ago
3 0

Answer:

Any Images?

Step-by-step explanation:

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If a batch of cookies uses
lianna [129]

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13 1/3 cups of sugar

Step-by-step explanation:

8 0
2 years ago
12x15y=4140; x+y= ? <br> solve.
Semenov [28]

Answer: The number of $12 tickets and $15 sold are 120 and 180 respectively.

Step-by-step explanation:

The given equation is:

12x + 15y = 4140             .....(1)

Where,

x stands for $12 tickets and y stands for $15 tickets

According to the question:

x + y = 300           ....(2)

Now solving equation 1 and 2.

From equation 2:

x = 300 - y

Now putting this expression in equation 1.

12x + 15y = 4140

12(300 - y) + 15y = 4140

3600 - 12y + 15y = 4140

3600 + 3y = 4140

3y = 4140 - 3600

3y = 540

y = 180

And,

x = 300 - y = 300 - 180 = 120

The number of $12 tickets sold = x = 120

The number of $15 tickets sold = y = 180

Thus, the number of $12 tickets and $15 sold are 120 and 180 respectively.

4 0
2 years ago
GIVING BRAINLIEST PLEASE HELP
BabaBlast [244]

Answer:

S=16×2

S=32

................................

3 0
3 years ago
Read 2 more answers
Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding no
baherus [9]

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.

5 0
3 years ago
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