(18 + 4i) + (-11 + 23i) =<br> Express your answer in the form (a + bi).

In the equation y=2x+1 what is true about the slope

ivolga24 [154]

The answer is 3 !!!!!!!!

5 0

3 years ago

What is the probability of randomly drawing an odd number between 24 and 33?

Studentka2010 [4]

Write out the numbers between 24 and 33: {24, 25, 26, 27, 28, 29, 30, 31, 32, 33}

How many numbers have we here? 10.

How many of these numbers are odd? {25, 27, 29, 31, 33}

Strictly speaking, "between 24 and 33" does not include {24, 33}.

Thus, the odd numbers between 24 and 33 are {25, 27, 29, 31}

The chances of drawing an odd number between 24 and 33 are then 4 / 10.

If, however, we omit the endpoints 24 and 33, then there are 8 numbers between 24 and 33: {25, 27, 29, 31}

and the odds of choosing an odd number from these eight numbers is 4/8, or 1/2, or 0.50.

6 0

3 years ago

In a clinical trial of 268 subjects treated with a drug, 11% of the subjects reported dizziness. The margin of error is plus o

Mariulka [41]

Answer:

$\hat p =0.11$ represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

$\hat q = 1-\hat p = 1-0.11= 0.89$ represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

$\alpha= 1-0.90 =0.1$ and this value represent the significance level of the test or the probability of error type I

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

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

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error is given by:

$ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

$Lower= 0.11-0.04 = 0.07$

$Upper= 0.11+0.04 = 0.15$

And for this case we have the following info:

$\hat p =0.11$ represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

$\hat q = 1-\hat p = 1-0.11= 0.89$ represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

$\alpha= 1-0.90 =0.1$ and this value represent the significance level of the test or the probability of error type I

4 0

3 years ago

How you solve this ?

jonny [76]

You have to use Pythagorean theorem

12^2+20^2 = x^2

sqrt(144*400) = x^2
sqrt(544) = x

X = 4 sqrt(34)

3 0

3 years ago

3x+10=4x+10 what is the answer?

ValentinkaMS [17]

Answer:

x is 0.

Step-by-step explanation:

Subtract both sides by 3x. Then, you will have 10= x+10. Subtract 10 on both sides to isolate x, then you will have x=0. Therefore, this equation on both sides has a solution of 0. Hope this helps!

4 0

3 years ago

Read 2 more answers

(18 + 4i) + (-11 + 23i) = Express your answer in the form (a + bi).