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Soloha48 [4]
3 years ago
9

Layla has 3/4 of a can of paint. She uses 2/3 of the paint for one wall in her bedroom. How much of the can did she use? PEASE A

NWSER!
Mathematics
1 answer:
maksim [4K]3 years ago
4 0

Answer:

8/12 of the can

Step-by-step explanation:

Make the denominator of both 12. Then 2/3 is 8/12.

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Mr. jones has two children. the older child is a girl. what is the probability that both children are girls?
allsm [11]
Since we already know that the older child is a girl (a 100% chance to be a girl), there must be a 50% chance that the other child is a girl. Therefore, the probability that both children are girls is 50%.
Hope this helps!
3 0
3 years ago
I need some help on this question please
artcher [175]

Answer:

  36, 32, 28, 24

Step-by-step explanation:

Fill in the values of n and do the arithmetic.

  a1 = 36 -4(1 -1) = 36

  a2 = 36 -4(2 -1) = 32

  a3 = 36 -4(3 -1) = 28

  a4 = 36 -4(4 -1) = 24

_____

You could recognize the formula as the specific case of the explicit formula for an arithmetic sequence with first term 36 and common difference -4. That tells you the second term is 36 -4 = 32, and each successive term is 4 less than the one before.

7 0
3 years ago
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astraxan [27]
For the volume of triangular prism, volume = 0.5xaxbxc
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4 0
3 years ago
Read 2 more answers
Wite a numerical expression for the below problems. Simplify the expressions.
ziro4ka [17]

Answer:

1) 9

2) 34

Step-by-step explanation:

For the first one, it would be

= (6×3)/2

= 18/2

= 9

The second one is,

= 36- (12/6)

= 36 - 2

= 34

4 0
2 years ago
Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six
baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.  

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

P(A|d) =\frac{P(A \cap d)}{P(d)}

P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}

P(A|d)= 0.3

Hence, the probability that a part was manufactured on machine A is 0.3

8 0
3 years ago
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