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3 years ago

What are the steps I need to solve this problem? 2x + 5 = 21

Mathematics

1 answer:

Vladimir [108]3 years ago

8 0

Step-by-step explanation:

subtract 5 under itself and 21 , then cross out the 5 and only subtract 21 and 5 which is 16 and then bring down 2x so now your equation will be 2x=5 so now your gonna divide 2 and 5

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If 8 feet of fencing costs $12.40 , how much does 36 feet of fencing cost?

Vesnalui [34]

Answer:

$55.80

Step-by-step explanation:

8, 16, 24, 32, 36

8=12.40, 16=24.80, 24=37.2, 32=49.6,

+4 makes it 36 so it's have of 8, so add 6.20 to the end and it's 55.80

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3 years ago

In the xy-plane, a line crosses the y-axis at the point (0, 3) and passes through the point (4, 5). Which is the equation of the

Gelneren [198K]

Answer:D

Step-by-step explanation:

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2 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

3 years ago

The sum of two numbers is 21. The larger number is 6 less than twice the smaller number. Find the two numbers.

never [62]

The sum of two numbers
x + y = 21
x = smaller number
y = larger number

Larger Number = y = (2x -6)
Substitute (2x - 6) for y
x + y = 21
x + (2x - 6) = 21
3x - 6 = 21
3x = 21 + 6
3x = 27
3x / 3 = 27 / 3
x = 9
Smaller number = x = 9
Find the larger number by inserting 9 into (2x - 6)
(2x - 6)
(2(9) - 6)
18 - 6
12

So
x = 9
y =12

Check our answer by inserting x = 9 and y = 12 into x + y = 21
x + y =21
9 + 12 = 21
21 = 21
So since we have 21 = 21 we know our answer is correct.

5 0

3 years ago

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir

AVprozaik [17]

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.

8 0

3 years ago