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anyanavicka [17]
3 years ago
15

What is the LCM of 7 and 13

Mathematics
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:

91 .

Step-by-step explanation:

The least common multiple (LCM) of two or more non-zero whole numbers is the smallest whole number that is divisible by each of those numbers. In other words, the LCM is the smallest number that all of the numbers divide into evenly.  Glad I could help!!

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Rewrite 20/7 as a mixed number
Neko [114]
20/ 7
7 goes into 20, 2 times oh whatever the answer is
2 and 6/7
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3 years ago
2. Show that every even number greater than 6 and less than 30 is the sum of two
Archy [21]

Answer:

I don't know

Step-by-step explanation:

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7 0
2 years ago
Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. The wag
Nutka1998 [239]

Answer: 1.303639

Step-by-step explanation:

The t-score for a level of confidence (1-\alpha) is given by :_

t_{(df,\alpha/2)}, where df is the degree of freedom and \alpha is the significance level.

Given : Level of significance : 1-\alpha:0.80

Then , significance level : \alpha: 1-0.80=0.20

Sample size : n=40

Then , the degree of freedom for t-distribution: df=n-1=40-1=39

Using the normal t-distribution table, we have

t_{(df,\alpha/2)}=t_{39,0.10}=1.303639

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639

8 0
3 years ago
What is the discriminant of 3x squared -10x= -2
artcher [175]
The discriminant of the equation 3 x^{2} -10x=-2 is 76.
I hope that helps.
3 0
2 years ago
University dean is interested in determining the proportion of students who receive some sort offinancial aid. rather than exami
vladimir2022 [97]
The confidence interval is
0.59\pm0.081

We first find p, our sample proportion.  118/200 = 0.59.

Next we find the z-score associated with this level of confidence:
Convert 98% to a decimal: 98% = 98/100 = 0.98
Subtract from 1:  1-0.98 = 0.02
Divide by 2:  0.02/2 = 0.01
Subtract from 1:  1-0.01 = 0.99

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 2.33.

The margin of error (ME) is given by
ME=z\sqrt{\frac{p(1-p)}{n}}
\\
\\=2.33\sqrt{\frac{0.59(1-0.59)}{200}}=2.33\sqrt{\frac{0.2419}{200}}\approx0.081

This gives us the confidence interval 
0.59\pm0.081
8 0
3 years ago
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