One number is five times another number. Their sum is 42. What are the two numbers? Answer ASAP

6th grade math ! Help me pleaseee

Maksim231197 [3]

Answer:

$9.99

Step-by-step explanation:

To find the cost of 1 pizza pie, divide the total cost by 5

49.95 ÷ 5 = $9.99

3 0

2 years ago

Read 2 more answers

2/x + 3/y = 13 ; 5/x - 4/y = -2

Tomtit [17]

Given that

(2/x)+(3/y) = 13--------(1)

(5/x)-(4/y) = -2 -------(2)

Put 1/x = a and 1/y = b then

2a + 3b = 13 ----------(3)

On multiplying with 5 then

10a +15 b = 65 -------(4)

and

5a -4b= -2 ----------(5)

On multiplying with 2 then

10 a - 8b = -4 -------(6)

On Subtracting (6) from (4) then

10a + 15b = 65

10a - 8b = -4

(-)

_____________

0 + 23 b = 69

______________

⇛ 23b = 69

⇛ b = 69/23

⇛ b =3

On Substituting the value of b in (5)

5a -4b= -2

⇛ 5a -4(3) = -2

⇛ 5a -12 = -2

⇛ 5a = -2+12

⇛ 5a = 10

⇛ a = 10/5

⇛ a = 2

Now we have

a = 2

⇛1/x = 2

⇛ x = 1/2

and

b = 3

⇛1/y = 3

⇛ y = 1/3

Answer :-The solution for the given problem is (1/2,1/3)

Check: If x = 1/2 and y = 1/3 then

LHS = (2/x)+(3/y)

= 2/(1/2)+3/(1/3)

= (2×2)+(3×3)

= 4+9

= 13

= RHS

LHS=RHS is true

and

LHS=(5/x)-(4/y)

⇛ 5/(1/2)- 4/(1/3)

⇛(5×2)-(4×3)

⇛ 10-12

⇛ -2

⇛RHS

LHS = RHS is true

5 0

2 years ago

The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample

dem82 [27]

The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. The margin of error of the sample mean is 83.71%.

7 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

We can conclude that the change to the face 3 affects the value of P(Odd Number)

7 0

2 years ago

It took 7.2 minutes to upload 8 digital photographs from your computer to a website. At this rate, how long will it take to uplo

kenny6666 [7]

If it take some x minutes to upload some y digital photographs, it means it has an average of: x/y minutes to upload a single digital photograph. Then, to upload z photographs, you just multiply x/y by z.

Spoilers for answer:
If it take 7.2 minutes to upload 8 digital photographs, it means it has an average of: 7.2/8 = 0.9 minutes to upload a single digital photograph. This means that, by this rate, it will take 20 * 0.9 = 18 minutes to upload 20 photographs to the website.

5 0

3 years ago