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3 years ago

lessils used 25% of half a gallon (64 ounces) of milk for a recipe how many ounces of milk where left

Mathematics

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

There were 48 ounces of milk left

Step-by-step explanation:

∵ Lessils used 25% of half a gallon of milk

→ Assume that the total amount of the milk is 25%

∴ The amount of milk left = 100% - 25% = 75%

∵ Half gallon = 64 ounces

→ That means the amount of milk left is 75% of 64 ounces

∴ The amount of milk left = 75% × 64

→ Divide 75% by 100 to change it to decimal

∵ 75% = 75 ÷ 100 = 0.75

∴ The amount of milk left = 0.75 × 64

∴ The amount of milk left = 48 ounces

∴ There were 48 ounces of milk left

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Assume that blood pressure readings are normally distributed with a mean of 123 and a standard deviation of 9.6. If 144 people a

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Step-by-step explanation:

In this question, we are asked to calculate the probability that the mean blood pressure readings of a group of people is less than a certain reading.

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3 years ago

If the hypotenus is 41 inches, and one of the legs is 40 inches what is the length of the other Leg

andre [41]

You find the answer through the Pythagorean theorem

a^2 + b^2 = c^2 a & b are the legs and c is the hypotenuse
a = 40
b = ?
c = 41

40^2 + b^2 = 41^2
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i hope you found this helpful!

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3 years ago

In a particular course, it was determined that only 70% of the students attend class on Fridays. From past data it was noted tha

siniylev [52]

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Step-by-step explanation:

Since we have given that

Probability that students attend class on Fridays = 70% = 0.7

Probability that who went to class on Fridays would pass the course = 95% = 0.95

Probability that who did not go to class on Fridays would passed the course = 10% = 0.10

Let A be the event students passed the course.

Let E be the event that students attend the class on Fridays.

Let F be the event that students who did not attend the class on Fridays.

Here, P(E) = 0.70 and P(F) = 1-0.70 = 0.30

P(A|E) = 0.95, P(A|F) = 0.10

We need to find the probability that they did not attend on Fridays.

We would use "Bayes theorem":

$P(F\mid A)=\dfrac{P(F).P(A\mid F)}{P(E).P(A\mid E)+P(F).P(A\mid F)}\\\\P(F\mid A)=\dfrac{0.30\times 0.10}{0.70\times 0.95+0.30\times 0.10}\\\\P(F\mid A)=\dfrac{0.03}{0.695}=0.043$

Hence, probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

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4 years ago