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agasfer [191]
4 years ago
8

73.93÷100 is what grade

Mathematics
2 answers:
eimsori [14]4 years ago
5 0
73.93 ÷ 100 =<span> 0.7393


Good Luck! </span>
uranmaximum [27]4 years ago
4 0
<span>Lets look at this:-

73.93 ÷ 100 = ? 
Lets solve:- 
</span><span>73.93 ÷ 100 = 0.7393
</span>
So, <span>73.93 ÷ 100 = 0.7393 and this is a C.

Hope i helped ya!! </span>
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Two groups of students were asked how far they lived from their school. The table below shows the distances in miles:
pychu [463]

Answer:

C. The value for Group A is double the value for Group B

Step-by-step explanation:

First, you add all the numbers from each group.

A: The distance in total is 39.06

B: The distance in total is 19.53

To find the mean, which is another way of saying the average, you divide the sum with how many numbers there are. (hope this makes sense, I suck at wording)

A: There are 9 numbers in total, hence, 39.06/9 = 4.34

B: There are 9 numbers in total, hence, 19.53/9 = 2.17

Those are the mean distance for the two groups.

2.17 * 2 = 4.34

Hence, C (third statement)

3 0
3 years ago
Read 2 more answers
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
4 years ago
Suppose 13% of American adults believe that the highest priority of America is to protect freedom of speech. We will take a rand
Cerrena [4.2K]

Answer:

A) neither Normal, not Binomial

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4 years ago
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7 0
2 years ago
Read 2 more answers
Answer numbers 16-18
lana66690 [7]
Number 16- 216
number 17- 180
number 18- 252
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3 years ago
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