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coldgirl [10]
3 years ago
8

John has 10 apples Sam has 20 apples how many apples are there in total

Mathematics
1 answer:
ratelena [41]3 years ago
6 0
Add 10 + 20 = 30 apples in total
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dan eat 3 boxes of jellybeans. if the re are 12 jellybeans in wach box , how mean jellybeans did dan eat in all
rjkz [21]
Multiply twelve by three and the ans is 36
8 0
4 years ago
Read 2 more answers
Write an equation you could use for this word problem. Do not need to solve.
blondinia [14]

Answer:

23 + x = 40

Step-by-step explanation:

Hi,

23 is the total amount your friend has. 23 + the total amount you have is equal to 40. Since we don't know how much you have, we can write x. Therefore, 23 + x = 40 would be correct.

Hope this helps :)

7 0
3 years ago
Read 2 more answers
6) Convert 2pie/3 to degrees<br> A) 30°<br> B) 60°<br> 120°<br> D) 240°
Gennadij [26K]
It would be c . 1.20
4 0
3 years ago
EASY MATH GOOD POINTS
34kurt

Answer:

t = (D/C - 1) (100/r)

Step-by-step explanation:

D = C(1+rt/100)

D/C = 1 + rt/100

D/C - 1 = rt/100

D/C - 1 = t (r/100)

Therfore,

<h2>t = (D/C - 1) / r/100</h2>

Therefore the last option is correct.

<h2><em><u>PLEASE MARK MY ANSWER AS BRAINLIEST!!!!!</u></em></h2>
8 0
4 years ago
Read 2 more answers
What percentage of babies born in the United States are classified as having a low birthweight (&lt;2500g)? explain how you got
lawyer [7]

Answer:

2.28% of babies born in the United States having a low birth weight.

Step-by-step explanation:

<u>The complete question is</u>: In the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g. What percent of babies born in the United States are classified as having a low birth weight (< 2,500 g)? Explain how you got your answer.

We are given that in the United States, birth weights of newborn babies are approximately normally distributed with a mean of μ = 3,500 g and a standard deviation of σ = 500 g.

Let X = <u><em>birth weights of newborn babies</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 3,500 g

            \sigma = standard deviation = 500 g

So, X ~ N(\mu=3500, \sigma^{2} = 500)

Now, the percent of babies born in the United States having a low birth weight is given by = P(X < 2500 mg)

         

   P(X < 2500 mg) = P( \frac{X-\mu}{\sigma} < \frac{2500-3500}{500} ) = P(Z < -2) = 1 - P(Z \leq 2)

                                                                 = 1 - 0.97725 = 0.02275 or 2.28%

The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.

4 0
4 years ago
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