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Valentin [98]
3 years ago
9

If

Mathematics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

5 + 2i

Step-by-step explanation:

We were given the complex numbers;

a =  - 2 - 5i

and

b =  - i

We want to find the product;

a {b}^{3}

We substitute the complex numbers into the expression and simplify:

( - 2 - 5i)( {i)}^{3}

This is rewritten as:

( - 2 - 5i)( {i)}^{2}  \times i

Note that

{i}^{2}  =  - 1

We substitute to obtain:

( - 2 - 5i)  \times - iLet us expand to get:

- 2 \times  - i + 5i \times  - i

This simplifies to:

2i - 5 {i}^{2}

This gives:

2i - 5( - 1) = 2i + 5

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Answer:

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Step-by-step explanation:

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find the value of r so that the line that passes through (5, r) and (2, -3) has a slope of 4/3. a. 1 b. 7 c. (13/9) d. -1
Margaret [11]
Slope  = (-3 - r) / (2 - 5) 

Slope = <span>(-3 - r) / -3 = 4/3

-(-3 - r) = 4

3 + r = 4

r = 4 - 3

r = 1   option b.</span>
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A radio DJ frequently announces how much money is currently in a jackpot. Every day several randomly selected residents are call
Fofino [41]

Answer:

"Variable interval" is the right solution.

Step-by-step explanation:

  • A variable-interval timetable seems to be a fiber-reinforced routine where another sensitivity or reaction would be commended because an unanticipated or unstable transaction has taken place, which would be the exact reverse of either a fixed-interval routine.
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Factories: 64x3y3+8-2xy
Elena-2011 [213]

can you make the question clear so I can answer you the question.

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3 years ago
The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
timurjin [86]

Answer:

a) Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

b) p_v =2*P(t_{(24)}  

c) 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 true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The values obtained are:

\bar X=85272 represent the mean annual salary for the sample  

s=11039.23 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =90000 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)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:  

Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t 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".  

Part b: Calculate the statistic

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

t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(24)}  

Part c: 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 true mean for the salary differs from 9000 at 5% of significance.

6 0
4 years ago
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