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lesya [120]
3 years ago
8

How does finding the square root of a number compare to finding the cube root of a number? Use the number 64 in your explanation

.
Mathematics
2 answers:
Jet001 [13]3 years ago
7 0

Answer:

Square is number which we get after multiply a number x to itself where as cube is a number which we get after multiplying a number x to itself three times.

Given number is 64

To find the square root or cube root of any number just right it using prime factorization method .

Then 64 can be written as 64=2×2×2×2×2×2

For square root we form <u>two pairs</u> of exactly same factors as 64=<u>2×2×2</u>×<u>2×2×2</u> = 8×8

∴ Square root of 64 =8

For cube root we form <u>three pairs</u> of exactly same factors as 64=<u>2×2</u>×<u>2×2</u>×<u>2×2</u>=4×4×4

∴ Cube root of 64 =4.


lakkis [162]3 years ago
3 0
In order to find the square root, we must find a number that when multiplied by itself one time, it produces the number under consideration.
In the case of 64, the square root is 8 because 8 multiplied once by itself is 64.

However, when finding the cube root, the root must multiply by itself two times and make the number. 
In the case of 64, 4 multiplied by itself two times is 64.
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A professor claims that his students' average score on the first exam of the semester is different than the average score on the
Marta_Voda [28]

Answer:

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value before (first exam) , y = test value after (second exam)

The system of hypothesis for this case are:

Null hypothesis: \mu_y -\mu_x =0

Alternative hypothesis: \mu_y -\mu_x \neq 0

The first step is calculate the difference d_i=y_i-x_i

The statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704

The next step is calculate the degrees of freedom given by:

df=n-1=8-1=7

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v = 2*P(t_{7} >0.704)=0.504

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

5 0
3 years ago
8/17 = y/7
STALIN [3.7K]

\frac{8}{17} = \frac{y}{7}

17y=56

y=3.3

4 0
3 years ago
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