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emmainna [20.7K]
3 years ago
12

Jane has a bag of candy, she gives 8 to her friend. Then jane's grandmother buys 5 more pieces than jane had originally. Now jan

e has 27 pieces of candy. How many pieces where originally in Jane's original bag of candy
Mathematics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

12 Pieces

Step-by-step explanation:

Number of candies in Jane's bag = x

she gives 8 to her friend;

x - 8

Then jane's grandmother buys 5 more pieces than jane had originally; add (x + 5)

x - 8 + x + 5

Now jane has 27 pieces of candy

x - 8 + x + 5 = 27

Solve for x;

x - 8 + x + 5 = 27

2x - 3 = 27

2x = 27 -3

2x = 24

x = 12

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Dominik [7]

Steps to solve:

4x + 1/3y^2 when x = 2 and y = 3

~Substitute

4(2) + 1/3(3)^2

~Simplify

8 + 1/3(9)

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8 + 3

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11

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6 0
3 years ago
What terms can be combined with 3a? Select all that apply.
antiseptic1488 [7]

Answer:

You can only add like terms with like terms, so the terms that can be combined with 3a are the terms that end in an a.

So, the answers are:

4a and 14a

Let me know if this helps!

6 0
3 years ago
Read 2 more answers
Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak
crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

P(0.47

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0
3 years ago
Simplify:<br> {[(16 ÷ 4) × (2 × 6)] ÷ 6} + 4 =
borishaifa [10]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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xz_007 [3.2K]

Answer:

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X = -16

Hope this helps

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Step-by-step explanation:

4 0
3 years ago
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