Alicia is a nurse.In 2016, she earned a monthly salary of £1750
In 2017, she was awarded a 2% pay increase on her monthly salary.
Given that in 2017 Alicia worked 45 hours per week for 48 weeks,
work out her average pay per hour for the year.
Answers
Final answer:
To calculate Alicia's average pay per hour for the year, we need to find her total earnings for the year and divide it by the total number of hours she worked.
Explanation:
To calculate Alicia's average pay per hour for the year, we need to find her total earnings for the year and divide it by the total number of hours she worked. In 2016, Alicia earned £1750 per month, so her annual salary would be £1750 x 12 = £21,000.
In 2017, she received a 2% pay increase, which means her monthly salary would be £1750 + 2% of £1750 = £1750 + (£1750 x 0.02) = £1750 + £35 = £1785.
Since Alicia worked 45 hours per week for 48 weeks in 2017, her total number of hours worked for the year would be 45 hours/week x 48 weeks = 2,160 hours. Now, we can calculate her average pay per hour by dividing her total earnings (£1785 x 12) by the total number of hours worked (2,160): £1785 x 12 / 2,160 = £10 per hour.
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AnswerAverage pay per hour 19.15
Step-by-step explanation:
1750 + 2% = 1785 pay increase
1785 x 12 months = 21420 a year
21420 divided by 52 weeks in a year averages to 411.92 weekly
Divide 411.92 by the 45 hours of work is 19.15 per hour
Brainliest please, if correct.
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Answer:
x=41.2
Step-by-step explanation:
hope this helps
Answer:
x=41.2
Step-by-step explanation:
First, subtract 7.8 from both sides of the equation, resulting in x-18=23.2
Then, add 18 to both sides of the equation, creating x=41.2
Hope this helps.
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Answer: There are 74 students of third grade attend an assembly .
Step-by-step explanation:
The number of students in third grade classes = 26
The number of third classes in the school = 3
Therefore, the total number of students of third grade in the school =
Now, it 4 students are absent then the number of third grade students in the school =
Hence, there are 74 students of third grade attend an assembly .
Answers
In our case independent variable is time in minutes
The dependent variable is the volume of air in the balloon.
b] The domain of the of this situation is:
[0≤t≤15]
this is because t will take 15 minutes for the balloon to get filled.
The range of the situation will be:
[0≤a(t)≤90000]
c]The function that represents this situation is a linear function given by:
y=ax+b
where:
a=rate/slope
b= initial amount of air in the balloon=0
Thus our function will be:
a(t)=6000t
d]
The total amount of air in the balloon after 6 minutes will be given by:
from
a(t)=6000t
a(6)=6000(6)
a(6)=36000 cubic feet
a. 7a-2b = 5a b
Answers
For this case we have the following expression:
From here, we must clear the value of a.
We then have the following steps:
Place the terms that depend on a on the same side of the equation:
Do common factor "a":
Clear the value of "a" by dividing the factor within the parenthesis:
Answer:
The clear expression for "a" is given by:
Answer: The required value of a is
Step-by-step explanation: We are given to solve the following equation for the value of a:
Since there are two unknowns and only one equation , so the value of a will definitely contain the value of b.
The solution of equation (i) for a is as follows:
Thus, the required value of a is
Answers
If it goes over the log, it simply travels the diameter, or 32 cm.
If it goes around the log, then it must travel half of the circumference, or 32*pi/2=16pi cm.
Note that this assumes that the inchworm and the raspberry patch are diametrically opposite.
Maybe I'm overthinking this one, but here's what I think it's talking about.
If I'm wrong, then I ought to at least get a few points for my talent at making
easy things difficult, and inventing obstacles to place in my own path.
-- The worm is 1 inch long.
-- The outside of the log is a cylinder. Its cross-section is a
perfect circle with a circumference of 32-cm.
-- The axis (length) of the log is perpendicular (across) the path
that leads to raspberry nirvana.
-- The ground is hard. The log contacts the ground along a line,
and doesn't sink into it at all.
-- The worm sees the log ahead of him. He continues crawling, until
he is directly under a point on the log that's 1-inch above him.
He then stands up to his full height, sticks his front legs to the log,
hoists himself up onto the bark, and starts to walk up and over it.
-- When he reaches a point on the other side of the log that's exactly 1-inch
above the ground, he hooks his sticky back feet to it, drops straight down to
the ground, and continues on his quest.
-- The question is: What's the length of the part of the log's circumference
that he traveled between the two points that are exactly 1-inch off the ground ?
I thought I was going to be able to be able to talk through this, but I can't.
I need a picture. Please see the attached picture.
Here comes the worm, heading from left to right.
He sees the log in front of him.
He doesn't bother going around it ... he knows he'll be able to get over it.
When he gets under the log, he starts standing straight up, trying to
grab onto the bark. But he can't reach it. He's too short, only 1 inch.
Finally, when he gets to point 'F', the bark is only 1" above him,
so he can hook on and haul himself up to point 'A'.
He continues on ... up, around, and over the log.
Eventually it dawns on him that the log won't last forever, and he'll
soon need to get down to the ground. As he comes down the right
side of the log, he starts looking down. It's too high. He can't reach
the ground, and he's afraid to jump.
Then he reaches point 'B'. It's exactly 1-inch above the ground, and
he leaves the log and gets down.
What was the length of the path he followed on the log ... the long way,
over the top from 'A' to 'B' ?
Here's what I did:
Draw radii from the center of the log to 'A' and 'B' .
Each of them is 16 cm long (1/2 of the diameter).
Draw the radius from the center of the log to the ground (' E ').
It's 16 cm all the way.
Point 'D' is 1 inch = 2.54 cm above the ground, so the
vertical leg of each little right triangle is (16 - 2.54) = 13.46 cm.
There are two similar right triangles, back to back, inside the log.
They are 'CAD' on the left, and 'CBD' on the right.
I want to know the size of the angles at the top of each triangle.
(One will be enough, since they're equal angles.)
For each of those angles, the side adjacent to it is 13.46 cm.
And the hypotenuse of each right triangle is a radius, so it's 16 cm.
The cosine of those angles is (adjacent/hypotenuse) = 13.46/16 = 0.84125 .
Each angle is 32.73 degrees.
Both of them put together add up to 65.45 degrees .
The full circumference of the log is (pi)(D) = 32pi cm.
The short arc between 'A' and 'B' is (65.45/360) of the full circumference.
The rest of the circumference is the distance that the worm crawled along it.
That's (1 - 65.45/360) times (32 pi) = (0.818) x (32 pi) = 82.25 cm .
Having already wasted enough time on this one in search of 5 points,
and then gone back through the whole thing to make corrections for
the customary worm crawling over the metric log, I'm not going to bother
looking for a way to check it.
That's my answer, and I'm sticking to it.
Answers
Number 12 have factors 12 = ( 1,2,3,4,6,12)
Number 18 have factors 18= ( 1,2,3,6,9,18 )
According to this we can conlude that numbers 12 and 18 have in common (1,2,3 and 6).
Good luck!!!
2 and 6 can go into it evenly