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Mars2501 [29]
3 years ago
5

Question 2 Rewrite in simplest radical form 1 x −3 6 . Show each step of your process.

Mathematics
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

√(x)

Step-by-step explanation:

(1)/(x^-(1/2)) that's 3 goes into -3 leaving 1 and goes into 6 leaving 2

1/2 is same as 2^-1

so therefore we can simplify the above as

x^-(-1/2)

x^(1/2)

and 4^(1/2)

is same as √(4)

so we conclude as

√(x)

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Assume that the growth rate of a population of ants is proportional to the size of the population at each instant of time. Suppo
Nutka1998 [239]
(1/A)  dA/dt=  C where A is the population of ants and C is a constant
ln(A) = C*t + C1 where C1 is another constant that comes out of integration and t is time in days.
Plugging in: at t=0, A= 100 so C1 = ln(100) = 4.605
at t=3, A=230 so ln(230) = 3*C +4.605 so C = 0.278
Final equation:
ln(A) = 0.278t + 4.605
or:
A = exp(0.278t + 4.605)
After 14 days, A = exp(0.278*14 + 4.605) = 4875.2
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3 years ago
Marking brianliest pls help
Contact [7]

Answer:

117.3

Step-by-step explanation:

7*4=117.29=117.3

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3 years ago
The graph shows the relationship between hours worked and total earnings for a store clerk. Let h represent hours worked and e r
Dovator [93]

Answer:

The answer is C.

Step-by-step explanation:

y = kx → e =

30

4

h → e = 7.5h

The independent variable is hours worked, h, and the dependent variable is total earnings, e. The independent variable is the input value of a function. The dependent variable is the output value of a function.

8 0
3 years ago
Identify the exponential function for the given graph.
Korolek [52]

Answer:

Step-by-step explanation:

C - 3(2^x).

8 0
3 years ago
The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, al
Valentin [98]

Answer:

a) p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation  

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \geq 0

Alternative hypothesis: \mu_y -\mu_x

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679

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

df=n-1=10-1=9

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

p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

Part b

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 confidence interval for the mean is given by the following formula:  

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that t_{\alpha/2}=3.25  

Now we have everything in order to replace into formula (1):  

-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469  

-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069  

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0
3 years ago
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