1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Liono4ka [1.6K]
3 years ago
13

a recipe uses 2 cups of milk per (1) serving. How many servings can be made by using 1 gallon of milk?

Mathematics
2 answers:
Scorpion4ik [409]3 years ago
3 0

Answer:

8

Step-by-step explanation:

16 cups per gallon divided by 2 equalls 8

Serhud [2]3 years ago
3 0

Answer:

8 servings

Step-by-step explanation:

1 serving uses 2 cups of milk, and there are 16 cups in a gallon.

you can use a 1:2 ratio to find that a gallon can make 8 servings.

You might be interested in
The student-to-faculty ratio at a small college is 17 to 3. The total number of students and faculty is 740. How many faculty me
ryzh [129]

17+3=20

740/20=37 Classes

37 classes x 3 faculty members=111

3 0
3 years ago
Pamela is 15 years younger than Jiri. The sum of their age is 33. what is Jiris age
Luden [163]

Answer:

24 years

Step-by-step explanation:

let jiri's age be x

Pamela's age=x-15

x+x-15=33

2x-15=33

collect the like terms to similar places

2x=33+15

2x=48

x=24

6 0
2 years ago
Determine the domain of the graph
ss7ja [257]

Answer:

The answer is B

Step-by-step explanation:

Graph goes positive infinitily,

but is inclusive of -4 and up.

7 0
3 years ago
Cual es la fracción de 0.7
kakasveta [241]

Answer:

La fracción de 0.7 es 7/10

8 0
3 years ago
The upper arm length of females over 20 years old in a country is approximately Normal with mean 35.8 centimeters (cm) and stand
lana [24]

Answer:

a) The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

b) 16% of women over 20 have upper arm lengths less than 33.3 cm.

Step-by-step explanation:

The Empirical Rule(68-95-99.7 Rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 35.8 cm

Standard deviation = 2.5 cm

(a) What range of lengths covers almost all (99.7%) of this distribution?

This range is from 3 standard deviations below the mean to three standard deviations above the mean.

So from 35.8 - 3*2.5 = 28.3 cm to 35.8 + 3*2.5 = 43.3 cm

The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

(b) What percent of women over 20 have upper arm lengths less than 33.3 cm?

68% of the women over 20 have upper arm length between 33.3 cm and 38.3 cm. The other 32% have upper arm length lower than 33.3 cm or higher than 38.3. The distribution is symmetric, so 16% of the have upper arm length lower than 33.3 cm and 16% have upper arm length higher than 38.3 cm

So 16% of women over 20 have upper arm lengths less than 33.3 cm.

4 0
3 years ago
Other questions:
  • How do you solve y=7x+35
    11·1 answer
  • How do i solve it pleas don't give the answer but give hints
    14·1 answer
  • A drink contains 20% cranberry juice and the rest is apple juice. What is the ratio of cranberry juice to apple juice? A.1:20 B.
    13·1 answer
  • Use the fractions 36in./1yd and 2.54 cm/1 in convert 3yd to cm
    9·1 answer
  • Which equations represent linear functions? Select all that apply.
    5·1 answer
  • Courtney needs to find 2,495 ÷ 48. In which place should she write the first digit of the quotient?
    12·2 answers
  • Which of the following is correctly written in Standard Form? −3x + 7y = 12, y = 3/7x + 6 ,5x − 4y = 9 ,3/7x + 2y =9
    7·1 answer
  • 5) Find the missing term in the sequence. What rule did you use?*
    9·1 answer
  • Solve for x:<br><br> x/8 - x/9 = 1
    12·1 answer
  • 3b - c +5 if c= - 4, b=2.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!