Coefficient of 24x+32+4x+3

Find the length x help pls , need asap

avanturin [10]

Answer:

x = 8

Step-by-step explanation:

$\frac{6}{x}$ = $\frac{3}{4}$

cross multiply

3x = 6(4)

3x = 24

divide

x = 8

3 0

3 years ago

Read 2 more answers

The FBI wants to determine the effectiveness of their 10 Most Wanted list. To do so, they need to find out the fraction of peopl

koban [17]

Answer:

98% of confidence intervals for the Population proportion of people who captured after appearing on the 10 most wanted list

(0.3583 , 0.4579)

Step-by-step explanation:

Explanation:-

Given sample size 'n' = 517

Given data Suppose a sample of 517 suspected criminals is drawn. Of these people, 211 were captured.

'x' =211

The sample proportion

$p^{-} = \frac{x}{n} = \frac{211}{517} =0.4081$

$q^{-} = 1-p^{-} = 1- 0.4081 = 0.5919$

98% of confidence intervals for the Population proportion of people who captured after appearing on the 10 most wanted list

$(p^{-} - Z_{\frac{\alpha }{2} } \sqrt{\frac{p^{-} (1-p^{-} )}{n} } , p^{-} + Z_{\frac{\alpha }{2} } \sqrt{\frac{p^{-} (1-p^{-} }{n} } )$

$(0.4081 - 2.326 \sqrt{\frac{0.4081 (0.5919 )}{517} } , 0.4081+ 2.326\sqrt{\frac{0.4081(0.5919 }{517} } )$

(0.4081-0.0498 , 0.4081 +0.0498)

(0.3583 , 0.4579)

Conclusion:-

98% of confidence intervals for the Population proportion of people who captured after appearing on the 10 most wanted list

(0.3583 , 0.4579)

7 0

4 years ago

Which expression can represent this numerical expression? 1/2+1/4

Ad libitum [116K]

Answer:

3/4

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

A 12-oz. jar of peanuts contains 8 servings. How many servings would a 32-oz. jar contain? Express your answer as a fraction or

Akimi4 [234]

Hope that helps!! Let me know if you don’t understand something :)

4 0

4 years ago

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