Simplify <br><br> 32 1/6

Solve x2 - 8x - 9 = 0.<br><br> Rewrite the equation so that it is of the form <br> x2 + bx = c.

stira [4]

X^2 - 8x-9=0
D = b^2- 4*a*c = 64- 4* (-9) = 64+36=100 =10^2
X1 = (8-10)/2 = -1
X2= (8+10)/2= 9

7 0

3 years ago

Justin is 2 years older than one-third Marissa's age Amy is 4 years older than two times Justin's age then find Justin's age and

Lana71 [14]

Yes I am pretty sure

6 0

3 years ago

Read 2 more answers

HELPPPP I NEED ASAP What region is it A,B,C,D

Maslowich

Region B

The solution of two intersecting lines is the point that both share in common. From a point of view, it is the point where the two lines meet. In this case, the point is in quadrant 1, or Region B.

8 0

3 years ago

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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

Use the box plots comparing the number of male puppies and number of female puppies attending doggy daycare each day for a month

dalvyx [7]

The IQR for the males' data is 25.
The difference between the median of the males' data and the female's data is 14.
The distribution of the males' data is skewed to the right and the median would be a better measure of the center. The distribution of the female's data is normal and the mean would be a better measure of the center.
A reason for the outlier is that the number of dogs needing care increased.

<h3>What is the interquartile range?</h3>

The interquartile range for the males data is the difference between the third quartile and the first quartile.

IQR = third quartile - first quartile

25 - 0 = 25

Median = 20 - 6 = 14

An outlier is a number that is way smaller or way larger than that of other numbers in a data set.

To learn more about outliers, please check: brainly.com/question/27197311

#SPJ1

7 0

2 years ago

Simplify 32 1/6 - 14 3/8