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Oxana [17]
3 years ago
6

Which question is a statistical question?

Mathematics
2 answers:
Alika [10]3 years ago
8 0

Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!

tinyurl.com/wpazsebu

jonny [76]3 years ago
6 0

Answer:

Step-by-step explanation:

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I need help with 13 on both pictures. The first picture is the answer and the second picture is the question not getting the ans
natita [175]

Problem 13

If we want to multiply (x^3-3x^2+2x) with (x^3-2x^2+x), then we can set up a diagram shown below. The terms are along the outside. The stuff inside is the result of multiplying each pair of outer terms.

  • Example: x^3 times x^3 = x^6 in the top left corner
  • Another example: 2x times x = 2x^2 in the bottom right corner.

This is known as the box method to keep track of all the terms multiplied.

Once the table is filled out, we add up each term inside the boxes. Combine like terms if possible. Notice that I color-coded the like terms (eg: the x^3 terms are in green boxes).

The final answer is x^6 - 5x^5 + 9x^4 - 7x^3 + 2x^2

6 0
2 years ago
Factor 2pq - 5qr + 10r - 4p.
Tresset [83]
The answer is B) <span>(q - 2)(2p - 5r) 

</span>2pq - 5qr + 10r - 4p = 2pq - 4p - 5qr + 10r
                                 = 2pq - 4p - (5qr - 10r)
                                 = 2p(q - 2) - 5r(q - 2)
                                 = (q - 2)(2p - 5r)
5 0
3 years ago
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
3 years ago
What times what give you 33
Tcecarenko [31]

Answer:

11x3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the volume of the cylinder
N76 [4]

Answer: wouldn't be 1608.5? let me know if im wrong.

V = πr^2 h= π · 8^2 · 8 ≈ 1608.49544

7 0
3 years ago
Read 2 more answers
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