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3 years ago

What is 6/8/ divided by 5/7

Mathematics

2 answers:

ziro4ka [17]3 years ago

8 0

Answer:

6/8 divided by 5/7 is 1.05/ 1 1/20

kondor19780726 [428]3 years ago

5 0

Step-by-step explanation:

do same change flip leave 6/8 the same then put × and flip 7/5 and it = 42/40 and u reduce it bye 2 =21/20 thats ur answer

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Look at the data set below:

Elenna [48]

Median: Sort from least to greatest and find the # in the middle.
Sort: 54, 58, 58, 60, 62.
Number in the middle: 58, so the median is 58.
Mode: 58, because it appears most.
Range: Subtracting bigger number from smaller number.
Finding it: 62-54 = 8
If you want to find the mean, then here:
Mean: Solving average (adding all #s together and dividing by how many numbers are there) <span>58 + 54 + 62 + 58 + 60 = 292.
</span>Divide by 5 = 58.4
Hope that helps you. Good luck with it.
Follow if you want.

6 0

3 years ago

Read 2 more answers

What times 18 gives you a factor of negative 12?

kozerog [31]

Well I personally think and this is just my opinion or should I say answer but I think it is 16

5 0

3 years ago

A researcher working in a Human Resources department was interested in gender and sales figures so he conducted a t-test. The me

irina1246 [14]

Answer:

$d = \frac{78.24 -66.25}{\sqrt{\frac{7^2 +7^2}{2}}}= 1.713$

For the interpretation we consider a value for d small is is between 0-0.2, medium if is between 0.2-0.8 and large if is higher than 0.8.

And on this case 1.713>0.8 so we have a large effect size

This value of d=1.713 are telling to us that the two groups differ by 1.713 standard deviation and we will have a significant difference between the two means.

Step-by-step explanation:

Previous concepts

The Effect size is a "quantitative measure of the magnitude of the experimenter effect. "

The Cohen's d effect size is given by the following formula:

$d = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1 +s^2_2}{2}}}$

Solution to the problem

And for this case we can assume:

$\bar X_1 =78.24$ the mean for females

$\bar X_2 =66.25$ the mean for males

$s_1 = s_2= 7$ represent the deviations for both groups

And if we replace we got:

$d = \frac{78.24 -66.25}{\sqrt{\frac{7^2 +7^2}{2}}}= 1.713$

For the interpretation we consider a value for d small is is between 0-0.2, medium if is between 0.2-0.8 and large if is higher than 0.8.

And on this case 1.713>0.8 so we have a large effect size

This value of d=1.713 are telling to us that the two groups differ by 1.713 standard deviation and we will have a significant difference between the two means.

4 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

M to the second power equals 49. Is it 7?

Feliz [49]

M^2 = 49
M^2 = 7^2
M = -7 and M = 7

4 0

3 years ago

Read 2 more answers