Answer: 
This is 17 to the 1/2 power
The rule is
![\sqrt[n]{x} = x^{1/n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B1%2Fn%7D)
the n is the index of the root. For square roots, n = 2, so,
![\sqrt[n]{x} = x^{1/n}\\\\\sqrt[2]{17} = 17^{1/2}\\\\\sqrt{17} = 17^{1/2}\\](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B1%2Fn%7D%5C%5C%5C%5C%5Csqrt%5B2%5D%7B17%7D%20%3D%2017%5E%7B1%2F2%7D%5C%5C%5C%5C%5Csqrt%7B17%7D%20%3D%2017%5E%7B1%2F2%7D%5C%5C)
Answer:
X =30/7
Step-by-step explanation:
The first way would be something like:
-7x + 10 = -20. The first thing we would do is to subtract 10 from both sides.
Therefore,
-7x+10-10=-20-10
Simplified, we get
-7x=-30
Remove the negative symbols
7x =30
x=30/7
Second example:
10=7x - 20
Add 20 to both sides.
10+20=7x-20+20
Simplified,
30=7x
x = 30/7
Ashton Martin. Because 60 divided by 20 is 3.
<h2>
Answer with explanation:</h2>
Let
be the average starting salary ( in dollars).
As per given , we have

Since
is left-tailed , so our test is a left-tailed test.
WE assume that the starting salary follows normal distribution .
Since population standard deviation is unknown and sample size is small so we use t-test.
Test statistic :
, where n= sample size ,
= sample mean , s = sample standard deviation.
Here , n= 15 ,
, s= 225
Then, 
Degree of freedom = n-1=14
The critical t-value for significance level α = 0.01 and degree of freedom 14 is 2.62.
Decision : Since the absolute calculated t-value (2.07) is less than the critical t-value., so we cannot reject the null hypothesis.
Conclusion : We do not have sufficient evidence at 1 % level of significance to support the claim that the average starting salary of the graduates is significantly less that $42,000.
11.25 cups of flour
2.5 x 4 = 10
2.5/2 = 1.25
10 + 1.25 = 11.25