(6 feets,8 inches) + (4 feets,10 inches)=? feets, ? inches

The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

AlladinOne [14]

We know for our problem that the zeroes of our quadratic equation are $(2-3i)$ and $(2+3i)$ , which means that the solutions for our equation are $x=2-3i$ and $x=2+3i$ . We are going to use those solutions to express our quadratic equation in the form $a x^{2} +bx+c$ ; to do that we will use the zero factor property in reverse:
 $x=2-3i$
$x-2=-3i$
$x-2+3i=0$

 $x=2+3i$
$x-2=3i$
$x-2-3i=0$

Now, we can multiply the left sides of our equations:
 $(x-2+3i)(x-2-3i)= x^{2} -2x-3ix-2x+4+6i+3ix-6i-3^2i^2$
= $x^{2}-4x+4-9i^{2}$
= $x^{2} -4x+4+9$
= $x^{2} -4x+13$
Now that we have our quadratic in the form $a x^{2} +bx+c$ , we can infer that $a=1$ and $b=-4$ ; therefore, we can conclude that $a+b=1+(-4)=1-4=-3$ .

6 0

3 years ago

Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card an

ziro4ka [17]

Answer:

1) is not possible

2) P(A∪B) = 0.7

3) 1- P(A∪B) =0.3

4) a) C=A∩B' and P(C)= 0.3

b) P(D)= 0.4

Step-by-step explanation:

1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4 . Thus the maximum possible value of P(A∩B) is 0.4

2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by

P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7

P(A∪B) = 0.7

3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3

4) the event C that the selected student has a visa card but not a MasterCard is given by C=A∩B' , where B' is the complement of B. Then

P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3

the probability for the event D=a student has exactly one of the cards is

P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4

3 0

3 years ago

What are the 4 translations for functions

sergiy2304 [10]

Rotation dilation translation reflection

8 0

3 years ago

Read 2 more answers

In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal

ddd [48]

Answer:

a) $z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

b) We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

Step-by-step explanation:

Previous concepts and data given

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".

$p_A$ represent the real population proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_A =\frac{26}{75}=0.347$ represent the estimated proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_A=75$ is the sample size for A

$p_B$ represent the real population proportion for women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_B =\frac{21}{64}=0.328$ represent the estimated proportion of women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_B=64$ is the sample size required for B

$z$ represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution

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

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:

Null hypothesis: $p_{A} = p_{B}$

Alternative hypothesis: $p_{A} \neq p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

Statistical decision

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

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

Part b

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.