All categories

3 years ago

Please give me the correct answer.Only answer if you're very good at math.Please don't put a link to a website.

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

3 0

Answer:

-79.78 (really stinking cold)

Step-by-step explanation:

Solving the equation will go like this:

5/9(F - 32) = -62.1

multiply both sides by 9 to get rid of the fraction

5 (F - 32) = -558.9

divide both sides by 5

F - 32 = -111.78

add 32 to both sides

F = -79.78

I checked the math twice.

Let me know if you have questions.

You might be interested in

1/4 divided by 2/3 show work pls

NARA [144]

$\dfrac{1}{4}:\dfrac{2}{3}=\dfrac{1}{4}\cdot\dfrac{3}{2}=\dfrac{1\cdot3}{4\cdot2}=\boxed{\frac{3}{8}}$

5 0

3 years ago

The slope of the line whose equation is given

rosijanka [135]

The slope is 9/1 hope i helped

8 0

3 years ago

Which is equivalent to -√10 3/4^x?

Galina-37 [17]

3√10^4x is equivalent to -√10 3/4^x and -4√10^3x they are all equivalent

4 0

3 years ago

Read 2 more answers

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 76.676.6

faltersainse [42]

Answer:

Step-by-step explanation:

Researchers measured the data speeds for a particular smartphone carrier at 50 airports.

The highest speed measured was 76.6 Mbps.

n= 50

X[bar]= 17.95

S= 23.39

a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?

If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps

b. How many standard deviations is that [the difference found in part (a)]?

To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation

Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations

c. Convert the carrier's highest data speed to a z score.

The value is X= 76.6

Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51

d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?

The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.

I hope it helps!

3 0

4 years ago

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago

Please give me the correct answer.Only answer if you're very good at math.Please don't put a link to a website.​

Please give me the correct answer.Only answer if you're very good at math.Please don't put a link to a website.