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ale4655 [162]
2 years ago
8

Determine the quotient of 3 over 7 divided by 2 over 3 . 6 over 21 1 over 2 9 over 14 1 and 5 over 9

Mathematics
1 answer:
Naddika [18.5K]2 years ago
8 1

um I dunno because I need points to ask a question sorry

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Can someone help me with this question please
Veronika [31]
The answer is 34 ft u have to apply pythogoras thereom
4 0
2 years ago
A company is designing a new cylindrical water bottle. The volume of the bottle will be 156156 cmcubed3. The height of the water
DaniilM [7]

Answer: the radius is 2.61 cm

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Volume = 156 cm³

Height = 7.3 cm

Therefore,

156 = 3.14 × r² × 7.3

156 = 22.922r²

r² = 156/22.922 = 6.81

Taking square root of both sides of the equation, it becomes

r = 2.61 cm

8 0
2 years ago
What is 865 divided by 40
Nutka1998 [239]
21.625 is your answer to 865/40
5 0
3 years ago
This rational expression is nonequivalent. 5x / 25x^4 ; x^3 / 5x^15<br> True or false?????
FromTheMoon [43]

It is nonequivalent. Figure it out this way. Think about what number has to be multiplied by both the numerator and the denominator of \frac{5x}{25x^4} to get to \frac{x^3}{5x^15}. It has to be the samee number for the one expression to be equivalent to the other. To get from 5x to x^3, we have to multiply by 1/5x^2. When we do that we get x^3. Good. Now we have to multiply the denominator by the same thing, 1/5x^2. 25x^4(\frac{1}{5}x^2)=5x^6. As you can see, they are not equivalent.

8 0
3 years ago
Read 2 more answers
An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-tempera
Olin [163]

Answer:

(115.2642, 222.7358).

Step-by-step explanation:

Given data:

type A: n_1=60, xbar_1=1827, s_1=168

type B: n_2=180, xbar_2=1658, s_2=225

n_1 = sample size 1, n_2= sample size 2

xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.

a=0.05, |Z(0.025)|=1.96 (from the  standard normal table)

So 95% CI is

(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]

=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)

= (115.2642, 222.7358).

8 0
3 years ago
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