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blagie [28]
3 years ago
5

The difference between two numbers is 16. Three times the larger number is seven times the smaller. What are the numbers?​

Mathematics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

12 and 28

Step-by-step explanation:

Represent the numbers using x and y.  Then y - x = 16.  (y is the larger one.)

Then 3y = 7x, and y = (7/3)x.

Substituting (7/3)x for y in y - x = 16, we get:

(7/3)x - x = 16.

Eliminate the fraction by multiplying all three terms by 3:

7x - 3x = 48, or 4x = 48, yielding x = 12.  Then y = (7/3)(12) = 28.

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Jake and Gloria are running on the track team. Gloria can run twice as many miles as
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Answer:

J+2J=12 so 3J=12 so J=4

Step-by-step explanation:

Use J to represent Jake's miles.

We know that Gloria's miles (G) is twice J, so G=2J

We also know that Jake and Gloria together ran 12 miles.

So we know, J+2J=12 or 3J=12   meaning J=4

So Jake runs 4 miles, and since Gloria is two times that she ran 8 miles.

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Solve the given inequality and graph the solution on a number line.
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Step-by-step explanation:

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If point C (3, 8) is reflected over the y axis what would be point C'?<br>no links please ​
olga2289 [7]

Answer:

(-3,8)

Step-by-step explanation:

When reflecting over the y-axis, you are switching from positive to negative x-values. Since 3 was already positive, we change the integer and in this case, to negative, so it becomes -3. Since the y value remains the same we don’t change the 8. So (-3,8)

I hope this helps!

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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.

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