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PtichkaEL [24]
4 years ago
9

What is the median of 6 and 13

Mathematics
1 answer:
solong [7]4 years ago
6 0
The median for 6 would be 7. The median for 13 would be 13
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HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!
tangare [24]

Answer: 8

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3 years ago
3 x 7/10 = ?
kap26 [50]

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a

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3 years ago
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Answer 2. Ella was born early in the morning on a Monday. She got married on the 9074th day of her life. What day of the week wa
m_a_m_a [10]

Answer:

Ella got married on a Wednesday.

Step-by-step explanation:

Let's solve this problem by understanding the following:

Each week is composed by 7 days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday.

So 1 week = 7 days;

Because the married day was on the 9074th day of her life, we can find the number of weeks that 9074 days represent:

\frac{1 week}{7 days} * 9074 days = 1296.285714 weeks

This means that 9074 days represent 1296.285714 weeks, which can be interpreted as 1296 entire weeks and a fraction of a week (0.285714).

Now let's calculate how many days 0.285714 weeks represent:

\frac{7 days}{1 week} * 0.285714 weeks = 2 days

This means that 9074 days are actually 1296 weeks and 2 days, because Ella was born on a Monday, and because after 7 days (1 week) it is Monday again, after 1296 weeks it is Monday, but as we also calculated 2 extra days, then the married day is two days after a Monday, that is a Wednesday.

In conclusion, Ella got married on a Wednesday.

3 0
3 years ago
An educational organization in California is interested in estimating the mean number of minutes per day that children between t
Hatshy [7]

Answer:

The critical value for a 98% CI is z=2.33.

The 98% confidence interval for the mean is (187.76, 194.84).

Step-by-step explanation:

We have to develop a 98% confidence interval for the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day.

We know the standard deveiation of the population (σ=21.5 min.).

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The z-value for a 98% CI is z=2.33, from the table of the standard normal distribution.

The margin of error is:

E=z\cdot \sigma/\sqrt{n}=2.33*21.5/\sqrt{200}=50.095/14.142=3.54

With this margin of error, we can calculate the lower and upper bounds of the CI:

LL=\bar x-z\cdot\sigma/\sqrt{n}=191.3-3.54=187.76\\\\\\UL=\bar x+z\cdot\sigma/\sqrt{n}=191.3+3.54=194.84

The 98% confidence interval for the mean is (187.76, 194.84).

4 0
3 years ago
How do you solve 5(3+k)=45
Helen [10]
We simplify the equation to the form, which is simple to understand
5(3+k)=45

Reorder the terms in parentheses
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Remove unnecessary parentheses
+15+5k=+45

We move all terms containing k to the left and all other terms to the right.
+5k=+45-15

We simplify left and right side of the equation.
+5k=+30

We divide both sides of the equation by 5 to get k.
k=6
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3 years ago
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