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Usimov [2.4K]
3 years ago
12

Can you answer the question in the picture pls :)

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
5 0
The answer is letter B
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Mrs. Portillo's cat weighs 6 kilograms. Her dog weighs 22 kilograms more than her cat. What is the total weight of her cat and d
topjm [15]
Cat=6
dog=22+6
total kilograms = 6+(22+6)= 34
5 0
3 years ago
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Please help me!!!!!! I really neeeeeeed
Butoxors [25]

Answer:

A-(x+4,y-5)

B- move 1 unit left and 7 units down

Step-by-step explanation:

A- 4 units right is x so

(x+4,y__)

5 units down is y so

(x+4,y-5)

B- move 1 unit left and 7 units down

because (x-1,y__) is going left because of the subtraction sign and its the x-coordinate, and (x-1,y-7) is down because its a Y-coordinate and sub so its going down.

<h2><u>TIP- Going left/down is always subtract/negative going right/up is positive/addition.</u></h2>
6 0
2 years ago
A store that buys and sells used video games buys a game at a cost of ​$9 and sells it at a selling price of ​$13. Find the mark
Bingel [31]
Markup = $4
b) markup as a percentage of cost is 33.3%
Step-by-step explanation:
Markup
markup = selling price - cost
= $13 - 9
... markup = $3
Markup as a Percentage of Cost
To find the percent markup, divide the markup by the reference value and multiply the ratio by 100%. The reference value for markup is usually cost price, but sometimes may be selling price.
... markup / cost × 100% = 3/9×100% = 33 1/3% ≈ 33.3%
4 0
3 years ago
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Exercise 6.13 presents the results of a poll evaluating support for the health care public option in 2009, reporting that 52% of
sleet_krkn [62]

Answer:

A sample size of 6755 or higher would be appropriate.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error M is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

52% of Independents in the sample opposed the public option.

This means that p = 0.52

If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Sample size of size n or higher when M = 0.01. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.01 = 1.645\sqrt{\frac{0.52*0.48}{n}}

0.01\sqrt{n} = 0.8218

\sqrt{n} = \frac{0.8218}{0.01}

\sqrt{n} = 82.18

\sqrt{n}^{2} = (82.18)^{2}

n = 6754.2

A sample size of 6755 or higher would be appropriate.

3 0
3 years ago
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storchak [24]
Use the Pythagorean Theorem to find the length of the line segment connecting (4,-6) and (9,-5).  The change in x is 5 and the change in y is 1.  Thus, the 
hypotenuse is     sqrt(5^2 + 1^2) = sqrt(26) (answer).
7 0
3 years ago
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