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Dmitriy789 [7]
2 years ago
15

The quotient of two less than a number and five is eight

Mathematics
1 answer:
postnew [5]2 years ago
5 0

Answer:

The number is 42.

Step-by-step explanation:

Let x be the unknown number.

"two less than a number" is x - 2

Quotient means the result of a division.

(x - 2)/5 = 8

Multiply both sides by 5.

x - 2 = 40

Add 2 to both sides.

x = 42

Answer: The number is 42.

Check.

Start with 42. 2 less than 42 is 40. The quotient of 40 and 5 is 40/5 which is 8. The answer is correct.

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WILL MARK BRAINIEST! 25 points!!! What is the x-coordinate of the solution to the following system of equations? 2x + y = −4 5x
vladimir2022 [97]

Answer:

x=-6

Step-by-step explanation:

2x+y=-4

5x+3y=-6

solve :

multiply first equation by 3

6x+3y=-12

5x+3y=-6

subtract the two equations : 6x+3y-5x-3y=-12-(-6)

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Answer:

(a) 860, (b) 860, (c) 860 and 186

Step-by-step explanation:

(a) 860

860 ends with a zero or a five, 186 and 863 do not.

(b) 860

860 ends with a zero, 186 and 863 do not.

(c) 860, 186

860, the 0 is even. 186, the 6 is even. 863, the 3 is not even.

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2 years ago
What is the force acting on a 16 kg object accelerating at 2.0 m/s2
alexira [117]

Answer:

32 n

Step-by-step explanation:

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6 0
2 years ago
Help show work please
pantera1 [17]
We want to get rid of the stuff in the denomenator

gcd of the denomenator is 2x
but wait, we can get rid of the deonmenator in the numerator by multiplying by 4x

muliplty the whole thing by  4x/4x

we get \frac{x(x^2)}{4(2)+2(x+4)}=\frac{x^3}{2x+16}
3 0
3 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
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