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UNO [17]
3 years ago
7

Insert three arithmetic means between 25 and 9​

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
6 0

Answer:

21, 17, 13

Step-by-step explanation:

Find the mean of the 2 given values

mean = \frac{25+9}{2} = \frac{34}{2} = 17

Noe find the mean of 25 and 17 and 17 and 9

mean = \frac{25+17}{2} = \frac{42}{2} = 21 and

mean = \frac{17+9}{2} = \frac{26}{2} = 13

Thus the 3 means are 21, 17 and 13, that is

25, 21, 17, 13, 9

You might be interested in
A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur
Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.1 = 0.9, so Z = 1.28.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28\frac{1.5}{\sqrt{305}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

8 0
3 years ago
Plz help me can't fail I need this and brainly is really helping
makvit [3.9K]

Answer:

1,2,3,4,5,6

odd numbers:

1,3,5

even numbers:

2,4,6

I would say its a 50/50 chance

4 0
2 years ago
Read 2 more answers
Please help me on this
timama [110]

Answer:

ANSWER IS C,,, 0

Step-by-step explanation:

the J+8 is basically going along with that the answer J+8 could be LESS than or equal to 8 if we pick any other number its going to be GREATER than 8 an we want LESS than 8,,,hope this helps.

8 0
2 years ago
I really really really need help!!!!
Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
8 0
2 years ago
Triangle HIJ has coordinates H(-3,5), (8,2), and J(-3,-5). Find
Vlad1618 [11]

Answer:

H (x= -3 , y=5)

I (x=8 , y=2)

J (x= -3 , y= -5)

D is the distance between the 2 points so just plug in the x and y cords of each of the corresponding points

so for (HI)

you plug in 2 on the red x

and 8 on the red y

then -3 in the blue x

and 5 in the blue y

5 0
2 years ago
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