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Olegator [25]
3 years ago
8

The appropriate mixing ratio for a solution to kill bloodborne pathogens is 1 part chlorine to 10 parts water. If you are mixing

a solution in a spray bottle with 15 ounces of water, how much chlorine do you need to add for the correct ratio?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
8 0
The ratio is:
Chlorine : Water = 1 : 10
m ( Water ) = 15 ounces
m ( Chlorine ) = x
x : 15 = 1 : 10
10 x = 15 * 1
10 x = 15
x = 15 : 10
x = 1.5 ounces
Answer:
We need to add 1.5 ounces of Chlorine for the correct ratio.
You might be interested in
What is the square root of 5 divided by the square root of 15 and simplify and in fraction form​
ozzi

Answer:

\sqrt{\frac{1}{3} }

or

\frac{\sqrt{3} }{3}

Step-by-step explanation:

The expression \frac{\sqrt{5} }{\sqrt{15} } can be simplified by first writing the fraction under one single radical instead of two.

\frac{\sqrt{5} }{\sqrt{15} } = \sqrt{\frac{5}{15} }

5/15 simplifies because both share the same factor 5.

It becomes \sqrt{\frac{5}{15} } = \sqrt{\frac{1}{3} }

This can simplify further by breaking apart the radical.

\sqrt{\frac{1}{3} }  = \frac{\sqrt{1} }{\sqrt{3} }  = \frac{1}{\sqrt{3} }

A radical cannot be left in the denominator, so rationalize it by multiplying by √3 to numerator and denominator.

\frac{1}{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} }  =\frac{\sqrt{3} }{3}

4 0
3 years ago
Which table shows ordered pairs that satisfy the function y=x^2+1
ohaa [14]

Where is the table? You could potentiallysolve this by deconstructing the function.

3 0
3 years ago
An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c
Helga [31]
2 times it will coincide

6 0
3 years ago
PLEASE HELP ME!!!!!!!!!!! WILL GIVE BRAINLIEST AND RATING!!!!!!!!!!!!
Kitty [74]
This will give us
2n-1-(-7n+2)=2n-1+7n-2=9n-3
That is the second option.
Please mark me as brainliest.
4 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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