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3 years ago

there are 25 servings on a 30.2 ounce jar of peanut butter . how many ounces of peanut butter are there in 1 serving

Mathematics

2 answers:

romanna [79]3 years ago

3 0

1.208 ounce because 30.2 divided by 25 equals 1.208 ounces

prohojiy [21]3 years ago

3 0

Answer: There are 1.208 ounces of peanut butter in one serving

Step-by-step explanation:

Hi, to answer this question we simply have to divide the total amount of peanut butter (30.2 ounces) by the number of servings (25).

Mathematically speaking:

30.2 ounces of peanut butter / 25 servings = 30.2/25=1.208

There are 1.208 ounces of peanut butter in one serving.

Feel free to ask for more if needed or if you did not understand something.

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Move the slider on the graph on the right to graph each function:

DedPeter [7]

Answer:

The domain is $x\geq 7$

The range is $y\geq 1$

Step-by-step explanation:

we have

$y=\sqrt{x-7} +1$

Find the domain

Remember that

The domain of a function is the set of all possible values of x

we know that the radicand of the function must be greater than or equal to zero

$x-7\geq 0$

solve for x

$x\geq 7$

therefore

The domain is $x\geq 7$

Find the range

Remember that

The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.

For x=7

The value of y is equal to

$y=\sqrt{7-7} +1=1$

The solution for y is the interval [1,∞)

therefore

The range is $y\geq 1$

6 0

3 years ago

Read 2 more answers

Solve the system of equations two ways: 3x+y=5 and x-2y=4 A) using substitution B) using elimination

Vedmedyk [2.9K]

A. solve for 1 variable
let's solve for x in 2nd equation
add 2y to both sides
x=2y+4
sub 2y+4 for x in other equation
3(2y+4)+y=5
6y+12+y=5
7y+12=5
minu12 both sides
7y=-7
divide 7
y=-1
sub back
x=2y+4
x=2(-1)+4
x=-2+4
x=2
(2,-1)

B. eliminate
eliminate y's
multiply first equation by 2 and add to first

6x+2y=10
x-2y=4 +
7x+0y=14

7x=14
divide by 7
x=2
sub back
x-2y=4
2-2y=4
minus 2
-2y=2
divide -2
y=-1
(2,-1)

(2,-1) is answer

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3 years ago

B is the midpoint of AC. AB = 2x+12 and BC = 5x + 10. Find AC

3241004551 [841]

Answer:

x=6

Step-by-step explanation:

/AB/=2x+12

/BC/=5x+10

3x+2=5x-10 subtract 2 from both sides

3x=5x-12 subtract 5x from both sides

-2x=-12 divide both sides by -2x

x =6

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2 years ago

a cone has a volume of 1230.88 units cubed and a diameter of 14 units what is the height of the cone use 3.14 for pi

ANEK [815]

Answer:

h= 24

Step-by-step explanation:

$v = \pi {r}^{2} \frac{h}{3}$

$r = \frac{d}{2}$

$r = \frac{14}{2}$

r= 7

1230.88=3.14×7^2×h/3

1230.88= 153.86h/3

h=1230.88/153.86

h= 24

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2 years ago

I NEED THIS ASAP, PLEASE HELP!!!

WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

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3 years ago