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morpeh [17]
2 years ago
15

The histogram to the right represents the weights​ (in pounds) of members of a certain​ high-school programming team. What is th

e class​ width? What are the approximate lower and upper class limits of the first​ class? 100 120 140 160 180 200 220 240 0 1 2 3 4 5 6 Weight (lbs) Frequency A histogram with horizontal axis labeled "Weight (pounds)" from 100 to 240 in intervals of 20 and vertical axis labeled "Frequency" from 0 to 6 in intervals of 1 contains seven adjacent vertical bars with heights as follows: 100 to 120, 4; 120 to 140, 1; 140 to 160, 1; 160 to 180, 3; 180 to 200, 3; 200 to 220, 4; 220 to 240, 2. What is the class​ width? The class width is nothing. ​(Simplify your​ answer.) What are the approximate lower and upper class limits of the first​ class? The approximate lower class limit is nothing. The approximate upper class limit is nothing. ​(Simplify your​ answers.)
Mathematics
1 answer:
AVprozaik [17]2 years ago
4 0

Answer:

Class width=20

Lower class limit of first class=100

Upper class limit of first class=120

Step-by-step explanation:

The class width can be calculated by taking the difference of two consecutive upper class limits or lower class limits.

Now we take any two consecutive upper class limits or lower class limits from classes 100-120,120-140,140-160,160-180,180-200,200-220,220-240.

We take upper class limits of first and second class i.e. 100 and 120.

Class width=120-100=20

Class width=20

The approximate lower and upper class limits of the first class from classes  100-120,120-140,140-160,160-180,180-200,200-220,220-240 are 100 and 120.

Class limits for first class is 100-120.

Lower class limit of first class=100

Upper class limit of first class=120

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marta [7]

Answer:

Step-by-step explanation:

Given is the probability distribution of a random variable X

X 4 5 6 7 Total

P 0.2 0.4 0.3 0.1 1

x*p 0.8 2 1.8 0.7 5.3

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a) E(X) = Mean of X = sum of xp = 5.3

Var(x) = 28.9-5.3^2=0.81

Std dev = square root of variance = 0.9

------------------------------------

b) For sample mean we have

Mean = 5.3

Variance = var(x)/n = \frac{0.81}{{36} } \\=0.0225

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3 0
2 years ago
Jay is letting her bread dough rise. After three hours, her bread dough is \dfrac{11}{5} 5 11 ​ start fraction, 11, divided by,
timama [110]

Answer:

Jay's bread size is 220% of the original size.

Step-by-step explanation:

The question is incomplete.

The complete question is as follows.

Jay is letting her bread rise. After 3 hours,her bread is at 11/5 of its original size. What percent of its original size is jays bread dough?

Solution:

Let the original bread size be = 100 units

After 3 hours the bread rises = \frac{11}{5} of the original size

New size of bread =  \frac{11}{5}\times 100=220 units

Percent of original size the new bread is

⇒ \frac{New\ bread\ size}{Original\ bread\ size}\times 100

⇒ \frac{220}{100}\times100

⇒ 220\%

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2 years ago
Tamara says that raising the number i to any integer power results in either -1 or 1 as the result, since i^2 = -1. Do you agree
Roman55 [17]
I agree only if you have even powers -- even negative ones.

1/i^2 = 1/-1 = - 1
i^0 also gives 1 So far no problem.

It is when you consider the odd numbers that you don't get 1 or -1 
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i^(4n + 1) =  i
i^(4n - 1) = -i

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2 years ago
Find the volume of this cylinder.
const2013 [10]

Volume of a Right circular cylinder=πr²h

Using "Cavaliers principle" or Just pire logic... it can be understood that...

if we were to cut of the base and top of the cylinder such that it will make up to be a right circular cylinder and join the other two pieces on top; We would obtain a Right circular cylinder With The same radius and height as the Given cylinder.

So;

Given. Base area B=81πcm²

also Base area =πr²

=>πr²=81π cm²

=>r=√81=9cm

Also Given;

Height h=3r

=>Height,h=3×9=27cm

So;Volume of cylinder V=πr²h ;(r²=81,h=27)

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Hope it helps...

Regards,

Leukonov/Olegion.

5 0
2 years ago
The graph of the function f(x) = (x – 4)(x + 1) is shown below. On a coordinate plane, a parabola opens up. It goes through (neg
DanielleElmas [232]

Answer:

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Step-by-step explanation:

we have

f(x)=(x-4)(x+1)

f(x)=x^2+x-4x-4

f(x)=x^2-3x-4

This is a vertical parabola open upward

The vertex is a minimum

The vertex is the point (1.5,-6.25)

we know that

The function is decreasing in the interval ----> (-∞,1.5)  x < 1.5

That means----> the function is decreasing for all real values of x less than 1.5

The function is increasing in the interval ----> (1.5,∞)  x> 1.5

That means----> the function is increasing for all real values of x greater than 1.5

see the attached figure to better understand the problem

therefore

The statement that is true is

The function is decreasing for all real values of x where x < 1.5.

5 0
2 years ago
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