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andrezito [222]
8 months ago
12

1/4 divided by 2/6 in simplest form

Mathematics
2 answers:
Nuetrik [128]8 months ago
3 0

Answer:

1/4*6/2=6/8

3/4

Step-by-step explanation:

VARVARA [1.3K]8 months ago
3 0
1/4*6/2=618
3/4
This is going to be your answer
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20.<br><br> What is the end behavior of the function?
Leokris [45]
A is the answer to your question
5 0
2 years ago
The volume of a cone of radius r and height h is given by V=πr²h³. If the radius and the height both increase at a constant rate
Aloiza [94]

Answer:

The volume of cone is increasing at a rate 1808.64 cubic cm per second.

Step-by-step explanation:

We are given the following in the question:

\dfrac{dr}{dt} = 12\text{ cm per sec}\\\\\dfrac{dh}{dt} = 12\text{ cm per sec}

Volume of cone =

V = \dfrac{1}{3}\pi r^2 h

where r is the radius and h is the height of the cone.

Instant height = 9 cm

Instant radius = 6 cm

Rate of change of volume =

\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{1}{3}\pi r^2 h)\\\\\dfrac{dV}{dt} = \dfrac{\pi}{3}(2r\dfrac{dr}{dt}h + r^2\dfrac{dh}{dt})

Putting values, we get,

\dfrac{dV}{dt} = \dfrac{\pi}{3}(2(6)(12)(9) + (6)^2(12))\\\\\dfrac{dV}{dt} =1808.64\text{ cubic cm per second}

Thus, the volume of cone is increasing at a rate 1808.64 cubic cm per second.

5 0
3 years ago
I am crying fo help<br>5.<br>6.<br>7.<br>8.
bixtya [17]
3*567=1701 because you have to muply
6 0
3 years ago
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A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a
natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

The sample means  = x₁  +  x₂  +  .   .   .  , x₂₃ / 23

= \frac{x_1 +x_2 +...x_2_3}{23}

= 2.3hr

⇒\sum xi =2.3 \times 23 = 52.9hrs

let x₁ , x₂, .   .   .  , x₂₃  arranged in ascending order

Then x₂₃ was 10  and has been changed to 14

i.e x₂₃ increase to 4

Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

therefore, the new sample mean is 2.47

2) For the old data set

the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = [{x_1_2]^t^h value remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0
3 years ago
Please help i’ll give brainliest please tysm :))
STALIN [3.7K]

Answer:

7

Step-by-step explanation:

Multiply the first collum by 7. It's basically unit rates

3 0
2 years ago
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