5x =2 writing your answer as a fraction in it’s simplest form

Question 5 Multiple Choice Worth 1 points) (04.05 MC) Determine if the ordered pair (-1 , -5) is a solution to the inequality y≤

Zielflug [23.3K]

Answer:

(-1, -5) satisfies the inequality and will be the solution to the inequality.

From the attached figure, it is clear that (-1, -5) is below the line.

Therefore, the correct option is:

''Yes, because 1-1, 5) is on the line''.

Step-by-step explanation:

Given the inequality

$y\le -\frac{3}{4}x-1$

substituting (-1, -5) to check whether it satisfies the inequality or not

$-5\le \:-\frac{3}{4}\left(-1\right)-1$

$-5\:\le \:\frac{3}{4}-1$

$-5\:\le -\frac{1}{4}$

TRUE

Thus, (-1, -5) satisfies the inequality and will be the solution to the inequality.

From the attached figure, it is clear that (-1, -5) is below the line.

Therefore, the correct option is:

''Yes, because 1-1, 5) is on the line''.

4 0

3 years ago

Find the value of x:

cluponka [151]

5x - 20 = 8x - 65
-20 = 3x - 65
3x = 45
x = 15

5 0

3 years ago

Read 2 more answers

3(у - 3) - 2(у 4) = 4

Dennis_Churaev [7]

Answer:

y = 5

Step-by-step explanation:

3(у - 3) - 2(у -4) = 4

Distribute

3y -9 -2y +8 =4

Combine like terms

y -1 =4

Add 1 to each side

y-1+1 = 4+1

y = 5

3 0

3 years ago

Read 2 more answers

Amal Worked a total of 30 hours last week. On Saturday And Sunday He Worked 5 times as many hours than he worked the rest of the

Lilit [14]

Let x represent the number of hours he worked during the weekdays (not Saturday or Sunday).
If x is how much he worked on the weekdays and he worked 5 times as much on Sat and Sun, then hopefully you agree that on Sat and Sun he worked 5x hours.
So we have 5x hours on the weekends and x hours on the weekdays, so in total for the whole week we have 5x + x = 6x hours in total.
The question tells us that he worked 30 hours total, so 6x = 30
Divide both sides by 6 to isolate x and we have x = 5.
He worked 5 hours the rest of the week.

Hope this helps. If it does, please be sure to make this the brainliest answer! :)

8 0

2 years ago

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.

Kobotan [32]

Answer:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=3.7$ represent the mean for athletes

$\bar X_{NA}=4.1$ represent the mean for non athletes

$s_{A}=1.1$ represent the sample standard deviation for athletes

$s_{NA}=1.3$ represent the sample standard deviation for non athletes

$n_{A}=32$ sample size for the group 2

$n_{NA}=45$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{NA}\geq 0$

Alternative hypothesis: $\mu_{A} - \mu_{NA}< 0$

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457$

P value

We need to find first the degrees of freedom given by:

$df=n_A +n_{NA}-2=32+45-2=75$

Since is a one left tailed test the p value would be:

$p_v =P(t_{75}$

Comparing the p value with a significance level for example $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.

5 0

3 years ago