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2 years ago

Last year Boris paid £256 for his car insurance.

Mathematics

1 answer:

jenyasd209 [6]2 years ago

6 0

Answer:

The percentage increase in the cost of Boris car insurance cost is 249%

Step-by-step explanation:

In this question, we want to calculate the percentage increase in the cost of car insurance paid by Boris

Mathematically, to calculate this percentage increase, we shall need to make use of a mathematical formula.

Mathematically, the percentage increase would be;

{(new value paid-old value paid)/old value paid} * 100%

From the question, we can identify that the old value paid is £256 while the new value paid is £894

Thus, the percentage increase would be ;

(894-256)/256 * 100% = 638/256 * 100 = 2.4921875 * 100 = 249.21875 which is 249% increase to the nearest whole percentage

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Ramon wants to plant cucumbers and tomatoes in his garden. He has room for

yarga [219]

Answer:

Number of Cucumbers = 12

Number of Tomatoes = 4

Step-by-step explanation:

Let number of cucumber be c and number of tomatoes be t

Since he has room for 16 plants, we can write:

c + t = 16

He wants to plant 3 times as many cucumbers as tomatoes. We can write:

c = 3t

We can substitute this in 1st equation and solve for t:

c + t = 16

3t + t = 16

4t = 16

t = 16/4 = 4

And c = 3t

c = 3(4) = 12

Number of Cucumbers = 12

Number of Tomatoes = 4

6 0

2 years ago

Read 2 more answers

Carlos and Maria drove a total of 233 miles in 4.4 hours. Carlos drove the first part of the trip and averaged 55 miles per hour

Mandarinka [93]

Let x - hours Carlos drove ; Let y - hours Maria drove

Equation 1: x + y = 4.4
x = 4.4 - y

Equation 2: 55x + 50y = 233
55(4.4 - y) + 50y = 233
242 - 55y + 50y = 233
-5y = 233-242
-5y = -9
y=9/5 or 1.8 hours Maria drove

To find time Carlos drove:
x=4.4 - 1.8
x = 2.6 hours

3 0

2 years ago

Read 2 more answers

gavmur [86]

Answer:

m∠QPM=43°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠NPQ=m∠MPN+m∠MPQ

we have

m∠NPQ=(9x-25)°

m∠MPN=(4x+12)°

m∠MPQ=(3x-5)°

substitute the given values and solve for x

(9x-25)°=(4x+12)°+(3x-5)°

(9x-25)°=(7x+7)°

9x-7x=25+7

2x=32

x=16

Find the measure of angle QPM

Remember that

m∠QPM=m∠MPQ

m∠MPQ=(3x-5)°

substitute the value of x

m∠MPQ=(3(16)-5)=43°

therefore

m∠QPM=43°

7 0

2 years ago

The segments shown below could form a triangle.

olga2289 [7]

By the Triangle Inequality Theorem, the sum of two sides should be greater than the length of the third side, while the difference of these two sides should be less than the length of this third side. Normally you would take the absolute value of the difference of these two side as you wouldn't know which is greater than the other!

The simplest way to prove whether these line segments can form a triangle, is by going against this theory. Let us prove that the line segment don't form a triangle. As you can see, adding 7 and 1 is greater than 1, respectively 7 and 7 is greater than 1. Thus -

Solution = A. True

3 0

1 year ago

Ben swims 50,000 yards per week in his practices. Given this amount of training, he will swim the 100-yard butterfly in 51.5

Elanso [62]

Answer:

MArginal productivity: $\frac{dt}{dL}=-0.0002$

We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

Step-by-step explanation:

The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.

In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.

The marginal productivity can be expressed as:

$\frac{dt}{dL}$

where dt is the variation in time and dL is the variation in training yards.

We can not derive the function because it is not defined, but we can approximate with the last two points given:

$\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002$

Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.

6 0

1 year ago