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2 years ago

What is the IQR? a. 20 b. 41 c. 11.5 d. 61

Mathematics

1 answer:

Andreas93 [3]2 years ago

8 0

Hello friend, (づ｡◕‿‿◕｡)づ I'm here to help you solve this

Answer:

Step-by-step explanation:

First Quartile Q1 = 15.75

Second Quartile Q2 = 30.5

Third Quartile Q3 = 51

Interquartile Range IQR = 35.25

Median = Q2 x˜ = 30.5

Minimum Min = 11.5

Maximum Max = 61

Range R = 49.5

~~Best regards

~Akmp

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Which expression is undefined? A. -8 divided by -8 B. -8 divided by 8 C. 0 divided by 8 D. 8 divided by 0

d1i1m1o1n [39]

Answer:

I'll go with A

Step-by-step explanation:

well -8 got no divisor so yeah

3 0

2 years ago

A prize fund is setup with a single investment of £4000 to provide an annual prize of £150. The fund accrues interest at 5% p.A.

Verdich [7]

The invested amount, A=£4000

Rate of interest on this amount, r=5% per annum.

Total interest on the amount £4000 after 1 year

$= 5\% \;\text{of} \;\£ 4000=0.05\times \£4000=\£200$ .

The total amount after 1 year, P=£4000+£200=£4200

Amount remains after an annual prize of £150=£4200-£150=£4050.

Note that after providing the annual prize the amount remains is £4050 which is more than the invested amount of £4000 as the prize money is less than the interest gained.

Now, for the 2nd year, the 5% interest is on the principal amount of £4050, which will have more interest as compared to the interest of 1st year.

Again, after prize distribution, the total amount remains for the principal amount of 3rd year will be more than the principal amount of 2nd year. This sequence repeated every year and the total amount will increase every year.

So, it will never end, the full prize money can be awarded for infinitely many years.

7 0

2 years ago

[25 POINTS] A cylinder has a volume of 400π cm3. Find the dimensions that minimize its surface area.

miss Akunina [59]

Answer:

r = 5.848cm, h = 11.696cm

Step-by-step explanation:

Surface area: $SA=2\pi rh+2\pi r^2$

Volume: $V=\pi r^2h$

Therefore, the height of a cylinder given its volume would be $h=\frac{V}{\pi r^2}$ , thus:

$SA=2\pi r(\frac{V}{\pi r^2})+2\pi r^2\\\\SA=\frac{2V}{r}+2\pi r^2\\\\SA=\frac{2(400\pi)}{r}+2\pi r^2\\\\SA=\frac{800\pi}{r}+2\pi r^2$

We now find the derivative of the surface area of the cylinder with respect to its radius and set it equal to 0, solving for the radius:

$\frac{d(SA)}{dr}=-\frac{800\pi}{r^2}+4\pi r$

$0=-\frac{800}{r^2}+4\pi r$

$0=800\pi+4\pi r^3$

$-800\pi=4\pi r^3$

$-200=r^3$

$r\approx5.848$

If $0If [tex]r>5.848$ , then the surface area of the cylinder increases

Therefore, the surface area is minimized when the radius is $r=5.848cm$ , making the minimum height $h=\frac{V}{\pi r^2}=\frac{400\pi}{\pi(5.848)^2}\approx11.696cm$ .