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3 years ago

Simplify i^7/i x i^3 A. i^4 B.i^11 C.i^3 D. 1

Mathematics

2 answers:

Liula [17]3 years ago

5 0

Answer:

c.i^ 3

Step-by-step explanation:

i^7/i x i^3

i ^ 7 / i ^ 4

i^ 7 - 4

i^ 3

mr_godi [17]3 years ago

5 0

I got -i in general, but it can be D

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What is the value of the coefficient of the fourth term of the expansion of (a+b)^5 ?

melomori [17]

Answer:

I don't know if you wanted it expanded using "Pascal's Triangle" or "Binomial Theorem" but either way is the same. Your expanded equation is:

a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

I hope this is any help to you, have a nice day.

Step-by-step explanation:

6 0

3 years ago

A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand

sesenic [268]

Answer:

The 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

Step-by-step explanation:

Previous concepts

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".

$\bar X_1 =80$ represent the sample mean 1

$\bar X_2 =75$ represent the sample mean 2

n1=25 represent the sample 1 size

n2=36 represent the sample 2 size

$\sigma_1 =5$ sample standard deviation for sample 1

$\sigma_2 =3$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =80-75=5$

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.88$

The standard error is given by the following formula:

$SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118$

Confidence interval

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

$5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898$

$5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102$

So on this case the 94% confidence interval would be given by $2.898 \leq \mu_1 -\mu_2 \leq 7.102$

7 0

3 years ago

The cost of a ticket increases by 10% to £18.15<br> Work out the original cost.

Licemer1 [7]

Answer:

£16.50

Step-by-step explanation:

$x \times (1+\frac{10}{100} )=18.15$

$x \times \frac{110}{100} =18.15$

$\frac{110}{100} x=18.15$

$1.1x=18.15$

$\frac{1.1x}{1.1}=\frac{18.15}{1.1}$

$x=16.5$

7 0

3 years ago

Read 2 more answers

On a map, the distance between two towns is 3.4 inches. The map scale is 2 inches: 25 miles. What is the actual distance between

Mariulka [41]

For this question we can solve it by setting up a proportion:

$\frac{2inches}{25 miles}= \frac{3.4inches}{x}$

Now we can solve for x to find the total distance between the towns. Let's cross multiply and solve for x:

$2x=85$
$x=42.5miles$

So now we know that the total distance between the towns is 42.5 miles.

4 0

3 years ago

Given f(x) = x + 5 and g(x) = 11x - 5, what is f(x) + g(x)?

juin [17]

F(x)+g(x)=(x+5)+(11x-5)=x+11x+5-5=12x using the commutative property.

6 0

3 years ago