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marusya05 [52]
3 years ago
13

Simplify the expression.

d="TexFormula1" title="\sqrt\frac{16}{15} - \frac{1}{3}" alt="\sqrt\frac{16}{15} - \frac{1}{3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
maxonik [38]3 years ago
5 0
Exact Form:
√15/4 -1/3

Decimal Form:
0.63491250

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/viewform?hr submission=Chclk4feKBIQ.
Gennadij [26K]

Answer:

c) .22

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

In this question:

\sigma = 2.5, n = 500

Then

M = z*\frac{\sigma}{\sqrt{n}}

M = 1.96*\frac{2.5}{\sqrt{500}}

M = 0.22

5 0
3 years ago
Ethan ran 11 miles in 2 hours. what is the unit rate of miles to an hour?
bearhunter [10]
11 ÷ 2 = 5.5
Ethan ran 5.5 miles per hour

4 0
3 years ago
Please Help Me!
alexdok [17]

Answer:

The answer to your question is below

Step-by-step explanation:

                 (2x⁴ + 7x³ - 3x + 9) - (6x⁴ - 3x³ - 6x² - 17)

Step 1. take out the parenthesis, in the second term, change the signs

                  2x⁴ + 7x³ - 3x + 9 - 6x⁴ + 3x³ + 6x² + 17

Step 2. Group like terms

                  (2x⁴ - 6x⁴) + (7x³ + 3x³) + (6x²) + (-3x) + (9 + 17)

Step 3. Simplify like terms

                          -2x⁴ + 10x³ + 6x² - 3x + 26

                 

3 0
3 years ago
What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of
notsponge [240]

Answer:

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, (x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)

<u>So, The coordinates of the triangle A'B'C' are:</u>

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

6 0
3 years ago
The drivers at Pizza Palace were loading up their 3 delivery vans for the evening. There were 65 pizzas. The boss told them to p
Lana71 [14]
65/3=21 with 2 leftover

(2/3)/3 or 2/3•1/3 = 2/9

so either they didn’t get to share or each got 2/9 of the leftovers
3 0
2 years ago
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