Here is an expression 3/5 divided by 1/4 ​Which situation matches this expression?
Which situation matches this expression?

A: An athlete runs 3/5 kilometers in 4 minutes. At that rate, how many kilometers does the athlete run in 1 minute?

B: A teacher fills an empty white pitcher with 3/5 gallons of water and an empty blue pitcher with 1/4 gallon of water. How many more gallons of water are now in the white pitcher than in the blue pitcher?

C: Mr. Smith has 3/5 pounds of beans. He uses 1/4 pound of beans each week. At the rate, for how many weeks will beans last?

D: A worker cuts pieces of string that are each 3/5 yards in length. How many pieces of string can the worker cut from a piece of string that is 1/4 yard in length?

Answer:

9.09

Step-by-step explanation:

[tex]\frac{0.81}{0.09} + \frac{0.81}{9}[/tex]

= 9 + [tex]\frac{0.81}{9}[/tex]

= 9+ 0.09

= 9.09

I hope this helped!! If so please rate and mark brainliest <3!!

Answer:

the answer to your question is B

Answer: X-axis

Step-by-step explanation: The arrow is pointing to the horizontal line on the on the graph. That horizontal line is the x-axis.

Answer:

50.29in

Step-by-step instructions:

First find the semi-circular part:- 1/2×2πr

=1/2×2×22/7×9

=28.29

Find whole perimeter:- 28.29+5+5+12 = 50.29 in

Answer:

1200

Step-by-step explanation:

since the first week has 75 doubling it is 150 doubling that is 300 doubling that is 600 doubling that for the final week is 1200.

1200 more zombies will there be after an additional 4 weeks.

A sequence is an enumerated group of items in mathematics where repetitions are permitted and order is important. It has members (also known as elements or phrases), just like a set. The length of the series is the number of elements (potentially infinite).

At initial the zombies are= 75

Now, the zombies doubles at every week.

So, After 1 st week the population of zombies are

= 75 x 2

= 150

Then, After 2 nd week the population of zombies are

= 150 x 2

= 300

Then, After 3 rd week the population of zombies are

= 300 x 2

= 600

Then, After 4th week the population of zombies are

= 300 x 2

= 1200

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Answer:

6.0 ft^2

Step-by-step explanation:

hope this helps

53x95 = 5035!

hopes this helps! ^^

Answer:

5605

Step-by-step explanation:

Answer:

She translated each vertex of Figure 1 seven units to the right.

Step-by-step explanation:

As per linear equation, there are 372 more girls in school than boys.

"A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is a variable, A is a coefficient and B is constant."

Given, there are 1970 pupils.

There are 799 boys in the school.

Therefore, the number of girls in the school is

[tex]= (1970-799)\\= 1171[/tex]

Therefore, the number of girls more than boys in the school is

[tex]= (1171-799)\\= 372[/tex]

In that school , there are 372 more girls than total number of the boys.

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The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

The amount of a radioactive substance remaining after t years is given by the function  

[tex]f(t)=m(0.5)^{t/h}[/tex]............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

[tex]f(10)=50(0.5)^{10/5.3}[/tex]

Therefore  the required equation  f(10) = 13.52 mg .

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Answer:

The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

What is the fromula for he amount of a radioactive substance remaining after t years?

The amount of a radioactive substance remaining after t years is given by the function  

............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

Therefore  the required equation  f(10) = 13.52 mg .

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What is it?

The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

How do you find IQR?

Step 1: Put the numbers in order. ...

Step 2: Find the median. ...

Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. ...

Step 4: Find Q1 and Q3. ...

Step 5: Subtract Q1 from Q3 to find the interquartile range.

Answer:

The answer to this quesion is a measure of statistical dispersion, the spread of data or observations, first you need to get the data find the median calculate the median from both lower and upper half of the data and the iqr is the diffrence between the lower and upper.

Step-by-step explanation:

I Hope This Helps You!

-Justin:)

Answer: 44 degrees

Step-by-step explanation:

The function is given by y = 550(0.76)ˣ, after 10 passes, the number of dirts is about 35.36 ml

Let y represent the amount of dirt on the floor after x passes.

The floor initially has 550 ml of dirt, hence a = 550.

During each pass, 24% of the dirt is removed. Hence b = 100% - 24% = 76% = 0.76

The equation is given by:

y = 550(0.76)ˣ

After 10 passes:

y = 550(0.76)¹⁰ = 35.36 ml

The function is given by y = 550(0.76)ˣ, after 10 passes, the number of dirts is about 35.36 ml

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Answer:

2 semi circles put together

Step-by-step explanation:

just cut the circle in half

first add the ratio numbers which is 12

[tex] \frac{2}{12} \times 6000 = 1000[/tex]

then divide 1000 with 175

=5.71

5.71×2.25

=12.85 ruro

Answer:

2x8=16 so there for y=x^18

Step-by-step explanation:

#1

First the rectangle

Area

For semicircle

Area

Total area

#2

Area:-

Answer:

20% of 30

= 20/100 * 30

= 6

hence 6 students recieved A grade

Step-by-step explanation:

Only 3 numbers are possible at units place (2, 4 & 6) as we need even number.

But at tens & hundreds place all 6 are possible

Number of 3 digit even numbers

= 3x6x6 = 108

36

[tex]Solution:[/tex]

[tex]6\times6=36[/tex]

[tex]such\ as:[/tex]

[tex]11\ \ \ 22\ \ \ 33\ \ \ 44\ \ \ 55\ \ \ 66\\ 12\ \ \ 21\ \ \ 31\ \ \ 41\ \ \ 51\ \ \ 61\\ 13\ \ \ 23\ \ \ 32\ \ \ 42\ \ \ 52\ \ \ 62\\ 14\ \ \ 24\ \ \ 34\ \ \ 43\ \ \ 53\ \ \ 63\\ 15\ \ \ 25\ \ \ 35\ \ \ 45\ \ \ 54\ \ \ 64\\ 16+263646+56\\ 6+6+6+6+6+6\\ =36.[/tex]

I hope this helps you

:)

Using the midline property;

[tex]2wy = vz[/tex]

[tex]2(p - 21) = p - 7[/tex]

[tex]2p - 42 = p - 7[/tex]

[tex]2p - p = - 7 + 42[/tex]

[tex]p = 35[/tex]

Answer:

its 54

Step-by-step explanation:

1+2=3 3x5=15 15x2=30 30+16=46 46+8=54

I know you're looking for an answer but i'm doing a challenge where i have to answer 10 questions by the end of today, I'm sorry! :(

Answer:

t = 0, and t = 4/5

Step-by-step explanation:

1. Write the equation in standard form:

-75t^2+60t=0

2. Solve with the Quadratic formula:

[tex]t_{1,2} = \frac{-60 +- \sqrt{60^2 - 4(-75)}0}{2(-75)} \\\\t_{1,2} = \frac{-60+-60}{2(-75)} \\\\3. Separate the fractions:\\t_{1} = \frac{-60+60}{2(-75)} , t_{2} = \frac{-60-60}{2(-75)} \\\\4. Solve for each:\\t_{1} = \frac{0}{-150} = 0\\t_{2} = \frac{-120}{-150} = 4/5[/tex]

hope this helps!

Answer:

Let a = side length of a cube

Let S = surface area of a cube

Area of a square = a²

Since a cube has 6 square sides:  S = 6a²

To make a the subject:

S = 6a²

Divide both sides by 6:

[tex]\sf \implies \dfrac{S}{6}=a^2[/tex]

Square root both sides:

[tex]\sf \implies a=\sqrt{\dfrac{S}{6}}[/tex]

(positive square root only as distance is positive)

-----------------------------------------------------------------------------------------------

[tex]\sf x=-3-\sqrt{2} \implies (x+[3+\sqrt{2}])=0[/tex]

[tex]\sf x=-3+\sqrt{2} \implies (x+[3-\sqrt{2}])=0[/tex]

Therefore,

[tex]\sf y=a(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]  for some constant a

Given the y-intercept is at (0, -5)

[tex]\sf \implies a(0+3+\sqrt{2}) (0+3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(3+\sqrt{2}) (3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(9-2)=-5[/tex]

[tex]\sf \implies 7a=-5[/tex]

[tex]\sf \implies a=-\dfrac57[/tex]

Substituting found value of a into the equation and simplifying:

[tex]\sf y=-\dfrac57(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]

[tex]\sf \implies y=-\dfrac57(x^2+6x+7)[/tex]

[tex]\sf \implies y=-\dfrac57x^2-\dfrac{30}{7}x-5[/tex]

Hey so I have a question do u need to explain?

Answer:

Third option

Step-by-step explanation:

Given the following question:

[tex]7x-4=5x+15[/tex]

To find the answer we need to use inverse of operations to simplify the equation in till we isolate the variable and find the value of x.

[tex]7x-4=5x+15[/tex]
[tex]-4+4=0[/tex]
[tex]15+4=19[/tex]
[tex]7x=5x+19[/tex]
[tex]5x-5x=0[/tex]
[tex]7x-5x=2x[/tex]
[tex]2x=19[/tex]
[tex]2x\div2=x[/tex]
[tex]19\div2=\frac{19}{2}[/tex]
[tex]=\frac{19}{2}[/tex]

Your answer is the third option or "19/2."

Hope this helps.

Answer:

Step-by-step explanation:

Notation for Cereal Set Roster

The roster form is one of the simplest ways to represent the set. The elements of a set are represented in a row and separated by a comma in roster form. For example, the first five positive even numbers are denoted as A = 2, 4, 6, 8, 10.

A = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100 is an example of roster notation: A = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.

Hope that helps you understand :)

Answer: 0.27

Population density of the species in water:

Population density of a species in the water is given by the expression,

       Density =Given in the question,

Number of goldfish in the tank = 12

Volume of the tank = 45 cubic feet

Substitute these values in the expression,

Density =

            = 0.267

            ≈ 0.27 goldfish per cubic feet

      Therefore, density of the tank will be 0.27 goldfish per cubic feet.