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Gnoma [55]
3 years ago
6

I don’t know if I’m correct or not help

Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer:

y=x+5

Step-by-step explanation:

you not subtracting, your adding five

the output is x+5 so whatever you input into the x, you add 5

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36 is 12% of what number?
Harman [31]
300


~

36 12%
--- = ---
X 100
8 0
3 years ago
Read 2 more answers
The manager of a large grocery store wants to inspect an incoming shipment of oranges to determine the
Nimfa-mama [501]

Answer:

b

Step-by-step explanation:

The variability of the proportion of blemishes for Mary's sample will be greater than the variability of Pat's.

7 0
3 years ago
In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

3 0
4 years ago
Mason opens a savings account by making a $165.85 deposit. Every week, he deposits another $20.50 in the account.
tiny-mole [99]

Answer:

i think that it would be

$370.85

Step-by-step explanation:

The following expression shows the amount of money in the account after x weeks.

If $20.50 was added to the account for 10 weeks we can substitue 10 as the x:

$370.85 will be in the account after 10 weeks

8 0
3 years ago
Read 2 more answers
Terry walked 3.5 kilometers over the weekend. Jamie walked 2.85 miles. Who walked more over the weekend? Explain your answer. (1
denis-greek [22]

Answer:

jamie.

Step-by-step explanation:

if 1 mile equals 1.6 km, then terry walked 2.1875 miles (3.5 divided by 1.6)

2.1875(what terry walked) is less than 2.85 miles. (what jamie walked)

6 0
4 years ago
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