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Scorpion4ik [409]
3 years ago
8

A teacher wants to find out the average number of weekly reading hours for her students. She noted the number of reading hours o

f 5 of her students, as shown below:
7, 8, 6, 5, 9

Which measure of central tendency is most appropriate for this situation and what is its value?
Mathematics
1 answer:
taurus [48]3 years ago
5 0

Since the numbers are close together and you do not have any outliers, the mean would be the best measure of central tendency.

The value is

(5+6+7+8+9)/5 = 7

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At a tea shop, a package of vanilla tea weighs 1 ounce and costs $2. Cinnamon tea is priced at 2 ounces per dollar. Use rates to
RoseWind [281]
The vanilla tea is more expensive
8 0
3 years ago
A small square cloth has an area of 361 cm2. Sarah sewed four of them together to form a larger square cloth. What is the perime
nika2105 [10]

Answer:

152 cm

Step-by-step explanation:

Once again its a square. All sides are equal.

Square root 361 to get 19.

19 x 19 = 361

4 was sewed together.

You could either add, 361+361+361+361

Or you could multiply, 361 x 4

Either way you get 1444.

You could either square root 1444 to get the side length or you could add the previous side length to itself once to get the new side length of 38.

Now you need to get perimeter.

P=4 x length

P=4 x 38

152 cm

4 0
4 years ago
Read 2 more answers
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
3 years ago
Whats is the square root of 2 multiply by one over the square root of 2
Ket [755]

When ANY number is multiplied by its own reciprocal, the product is ' 1 '.

8 0
4 years ago
Geometry
satela [25.4K]

Answer:

Option (C)

Step-by-step explanation:

From the picture attached,

Two lines 'h' and 'k' are the parallel lines and a transversal 'n' is intersecting these parallel lines at two distinct points.

By the theorem of alternate interior angles, angle 8 and angle 11 will be equal in measure.

∠8 ≅ ∠11 [Alternate interior angles]

m(∠8) = m(∠11) = 118°

Therefore, Option (C) will be the answer.

7 0
3 years ago
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