HELPP ME PLZ WILLING TO SEND MONEY! It’s urgent Davon needs 90 mL of a
10% acid solution for a science experiment. He has available a 5% solution and a 20% solution. How many millimeters of the 5% solution and how many millimeters of the 20% solution should he mix to make the 10% solution?

I think it would be 5 and 5 of each solution

Devon needs to add 60 mL of 5% acid solution and 30 mL of 20% acid solution

Step-by-step explanation:

It knows acid and water concentration for every solution. 2 equations are proposed because there are two unknowns. V1= volume 5%acid and V2 =volume 20% acid

Acid V1(0.05)+V2(0.2)=90mL(0.1)

Water V1(0.95)+V2(0.8)=90mL(0.9)

Clearing V1 from acid equation having to: V1=180-4V2

Replacing V1 in the water equation having to: (180-4V2)(095)+V2(0.8)=81

Clearing V2 = 30mL

Replacing in water equation V1(0.05)+30(0.2)=9 So, V1=60mL

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;