Graph the line by plotting any two ordered pairs that satisfy the equation.<br> -6y = 9

12 less than a number is -75 find the number

Minchanka [31]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Please help will give brainliest if you give proper answer

atroni [7]

On IV the answer is C

3 0

3 years ago

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In preparing a certain ointment, a pharmacist used 28.35 g of zinc oxide instead of the 31.1 g called for. Calculate the percent

Vilka [71]

Answer: 9.70%

Step-by-step explanation:

The formula to find the percentage error is given by :-

$\%\text{ error}=\dfrac{|\text{Estimate-Actual}|}{\text{Actual}}\times100$

Given : Actual mass of zinc oxide used by pharmacist = 28.35 g

Estimated mass of zinc oxide used by pharmacist =31.1 g

Now, $\%\text{ error}=\dfrac{|31.1-28.35|}{28.35}\times100$

i.e. $\%\text{ error}=\dfrac{2.75}{28.35}\times100$

i.e. $\%\text{ error}=9.70017636684\approx9.70\%$

Hence, the percentage of error on the basis of the desired quantity.= 9.70%

5 0

3 years ago

Explain how you would solve 5x - 3 = 9. (there should be a minimum of 2<br> complete sentences). *

SpyIntel [72]

Haha, you first start off with adding 3 to both sides, cancelling the -3, and adding it to 9. you end up with 5x=12, in which you divide both sides by 5. you get 2 and 2/5 or 12/5! hope that helps :)

7 0

3 years ago

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It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i

USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 55, \pi = \frac{24}{55} = 0.4364$

93% confidence level

So $\alpha = 0.07$ , z is the value of Z that has a pvalue of $1 - \frac{0.07}{2} = 0.965$ , so $Z = 1.81$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574$

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0

3 years ago

Graph the line by plotting any two ordered pairs that satisfy the equation. -6y = 9