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3 years ago

If it costs 3.99 for 24 ounces how much woud it cost for 1

Mathematics

2 answers:

MariettaO [177]3 years ago

5 0

Answer:0.16625

Step-by-step explanation:

asambeis [7]3 years ago

4 0

Answer:

~6.02

Step-by-step explanation:

It's a ratio

3.99 : 24

1 : 6.015037593984962

or....

1 : 6.02

I got my answer as "6.015037593984962" because from 3.99 to 1, you need to divide by 3.99 (any number divided by itself gives you 1), and so when you do something to one side, you do it to the other. So then I did 24 divided by 3.99 and i got "6.015037593984962". Since that number is big, I suggested rounding it up to the nearest hundredth, and that is at the number "1". Since the next number after "1" is "5" I rounded up "1" to "2" because....4 or less, let it rest, 5 or more, let it sore. And since it's the number 5, I rounded up the 1 to 2. And so I converted "6.015037593984962" to "6.02".

Hope this helped!

You might be interested in

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

8 0

2 years ago

A particular state has elected both a governor and a senator. Let A be the event that a randomly selected voter has a favorable

Elodia [21]

Answer:

Step-by-step explanation:

Given that A be the event that a randomly selected voter has a favorable view of a certain party’s senatorial candidate, and let B be the corresponding event for that party’s gubernatorial candidate.

Suppose that

P(A′) = .44, P(B′) = .57, and P(A ⋃ B) = .68

From the above we can find out

P(A) = $1-0.44 = 0.56$

P(B) = $1-0.57 = 0.43$

P(AUB) = 0.68 =

$0.56+0.43-P(A\bigcap B)\\P(A\bigcap B)=0.30$

a) the probability that a randomly selected voter has a favorable view of both candidates=P(AB) = 0.30

b) the probability that a randomly selected voter has a favorable view of exactly one of these candidates

= P(A)-P(AB)+P(B)-P(AB)

$=0.99-0.30-0.30\\=0.39$

c) the probability that a randomly selected voter has an unfavorable view of at least one of these candidates

=P(A'UB') = P(AB)'

= $1-0.30\\=0.70$

3 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

Starting with x=4.62 repeating express x as a fraction in simplest form

arsen [322]

$x=4.62=4+0.62=4+\dfrac{62}{100}=4+\dfrac{62:2}{100:2}=4+\dfrac{31}{50}\\\\\text{as a mixed number}\ 4\dfrac{31}{50}\\\\\text{as an improper fraction}\ \dfrac{4\cdot50+31}{50}=\dfrac{231}{50}$

5 0

3 years ago

I have four times as much money as my brother has. Our total is $55. How much does each of us have?

padilas [110]

Answer:

27.5

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers