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olasank [31]
3 years ago
15

How is the graph of y=^3√0.5x related to its parent function,y=^3√x ? A.It is horizontally stretched by a factor of 0.5 B. It is

horizontally compressed by a factor of 0.5 C. It is translated left by 0.5 units D. It is translated right by 0.5 units
Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0
For this case we have the following function:
 y = ^ 3√x
 Applying the following transformation we have:
 Expansions and horizontal compressions
 The graph of y = f (bx):
 If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.
 y = ^ 3√0.5x
 Answer:
 A.It is horizontally stretched by a factor of 1 / 0.5
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About 70% of Earth’s surface is covered by water. If the diameter of Earth is about 12,750 km, find the area not covered by wate
hichkok12 [17]
Earth diameter = 12,750 km
Earth radius = 6,375 km
Sphere Surface Area     =    <span> 4 • <span>π <span>• r²
</span></span></span>Sphere Surface Area     =    4 * 3.141<span>59 * 6375^2
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<span> <span> <span> Earth's Entire </span></span></span><span>Surface Area     =    510,704,724 square kilometers
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4 0
3 years ago
Can someone help please?
natima [27]
Question 4 is C and im not sure about the other questions
8 0
3 years ago
Evaluate, leave ur answer in index form. 2³ × 2^-1​
vitfil [10]

Multiply = add powers

3 + - 1 = 2

- - means +

+ + means +

- + means -

Thus, answer is 2^2

Hope this helps!

3 0
2 years ago
To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin
Murrr4er [49]

Answer:

a) t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

p_v =2*P(t_{15}>1.733)=0.104  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

b) t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

p_v =2*P(t_{15}>2.6)=0.0201  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation  

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

\bar X=31.3 represent the sample mean  

s=3 represent the sample standard deviation  

n=16 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 30, the system of hypothesis are :  

Null hypothesis:\mu = 30  

Alternative hypothesis:\mu \neq 30  

Since we don't know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>1.733)=0.104  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>2.6)=0.0201  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the we have a significant effect for the new treatment at 5% of significance.  

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0
4 years ago
Hour friends earned more than $400
borishaifa [10]

Answer:

the answer is D :)

Step-by-step explanation:

6 0
3 years ago
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