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wariber [46]
3 years ago
13

Two numbers with a total between 300 and 400​

Mathematics
1 answer:
Leokris [45]3 years ago
3 0
234 abd 127
It equals 361
If we add together different numbers lime 234 and 184 it would make 418 and thats too high.
Hope this helps:)
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Why do ⅚, 15,18, and 10/12 equal each other?
Burka [1]
10/12 is an un simplified form of 5/6.
8 0
3 years ago
Suppose the ages of multiple birth (3 or more babies) are normally distributed with a mean age of 31.7 years and a standard devi
artcher [175]

Answer:

The percent of these mothers are between the ages 30-35 is 36.53%

Step-by-step explanation:

we are given

mean of age =31.7 years

\mu=31.7

standard deviation of 5.2 years

\sigma=5.2

For age=30 years:

x=30

we can find z-score

z=\frac{x-\mu}{\sigma}

we can plug values

z=\frac{30-31.7}{5.2}

z=-0.32692

For age=35 years:

x=35

we can find z-score

z=\frac{x-\mu}{\sigma}

we can plug values

z=\frac{35-31.7}{5.2}

z=0.63462

now, we can use normal distribution table

P(-0.32692

now, we can find percentage

=0.3653\times 100

=36.53%


4 0
3 years ago
If the 1st step in the solution of the equation -9 + X = 5X -7 is "subtract X" then what should the next step be?
Fittoniya [83]

Answer:

the answer should be

Step-by-step explanation:

2nd step. ......-9+7=5x-x

3rd step............-2=4x

4th step.............-2/4=x

5th step.............-0.5=x

5 0
4 years ago
Billy has 24 feet of yarn how many inches of yarn does he have
EleoNora [17]
12 x 24 = 288 would be the answer
7 0
4 years ago
For a normal distribution, is it likely that a data value selected at random is more than 2 standard deviations above the mean?
makkiz [27]

Answer:

P(X>\mu + 2\sigma)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

Using the z score formula we want this probability:

P(Z>2)

And using the complement rule we got:

P(Z>2) = 1-P(Z

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interst of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

Where \mu the mean and \sigma  the deviation

We are interested on this probability

P(X>\mu + 2\sigma)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

Using the z score formula we want this probability:

P(Z>2)

And using the complement rule we got:

P(Z>2) = 1-P(Z

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

8 0
4 years ago
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