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kykrilka [37]
2 years ago
15

Translate each expression

Mathematics
1 answer:
melomori [17]2 years ago
4 0

Answer:

  • <em>Solving the first 6</em>
  • <em>Assumed the number is x</em>
<h3>#1</h3>
  • 4x - 1
<h3>#2</h3>
  • (2/3)x + 7
<h3>#3</h3>
  • m - n
<h3>#4</h3>
  • x² - 9
<h3>#5</h3>
  • 2x/5
<h3>#6</h3>
  • (1/4)x + 27
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The expression 28 + 63 is equivalent to<br> 1 91<br> 2 5,7<br> 3 6/7<br> 4 13,7
Yuliya22 [10]

Answer:

1) 91

Step-by-step explanation:

28 + 63 = 91

3 0
2 years ago
An amusement park reduced it's admission price to $15.50 per day, but now charges $1.50 per ride. Mark has $26 to spend on admis
gtnhenbr [62]
15.50+1.50x =26
1.50x = 10.50
x = 7
answer: he can ride 7 in one day
4 0
3 years ago
Maya shaved her head and then began letting her hair grow. She represents the length l of her hair, in
Drupady [299]
B

For example if it has been 2 months (l=1.25 times 2) the length would be 1.5 cm
3 0
3 years ago
A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken
mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

X=1 number of citrus trees that are infected with the citrus red mite.

n=5 random sample taken

\hat p=\frac{1}{5}=0.2 estimated proportion of citrus trees that are infected with the citrus red mite.

p true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56  

And rounded up we have that n=6147

5 0
3 years ago
Help me with hanger math...
andrew11 [14]
The answer is 3 First you subtract 2 from 17 and then you divide 15 by 5 to get the answer of 3
6 0
3 years ago
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