All categories

3 years ago

Decrease £28 by 7% Increase £28 by 7%

Mathematics

2 answers:

Shalnov [3]3 years ago

5 0

Answer:

decrease = 26.04

increase = 29.96

Step-by-step explanation:

Decrease of £28.00 by 7% = £28 -(7% of 28)

= £28 - (0.07 × 28)

= £28 - 1.96

= £26.04

Increase of £28 by 7% = £28 +(7% of 28)

= £28 +(0.7 × 28)

= $28 + 1.96

= £29.96

Decrease £28 by 7% would be £26.04

and increase £28 by 7% would be £29.96

Mrrafil [7]3 years ago

3 0

Increase : 29.96
Decrease: 26.04

You might be interested in

Normal Distribution Problem. Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy

GarryVolchara [31]

Answer: a) 0.9332, b) 0.8944

Step-by-step explanation: the probability value attached to a z score is gotten by using a z distribution table.

The z score is gotten by making use of the formulae below

Z = x - u / σ

Where x = sample mean, u = population mean and σ = population standard deviation.

For our question, u = 38, σ = 12, we are to look for the z score at z ≤ 56, that's x = 56

By substituting the parameters, we have that

z ≤ 56 = 56 - 38/ 12

z ≤ 56 = 18/12

z ≤ 56 = 1.5

To get the probabilistic value, we check the normal distribution table.

The table I'm using will be giving me probabilistic value towards the left of the area.

From the table, p ( z ≤ 56) = 0.9332.

Z >23 = 23 - 38/ 12

Z >23 = - 15/ 12

Z >23 = - 1.25

The probability value of this z score is towards the right of the distribution but the table I'm using is only giving probability values towards the left.

Hence Z >23 = 1 - Z<23

From the table, Z<23 = 0.1056.

Z >23 = 1 - 0.1056

Z >23 = 0.8944

4 0

2 years ago

Read 2 more answers

Write an equation of the line that is perpendicular to 5x+20y=10 and passes through the point (8,3)

natali 33 [55]

Y = 32x - 29

You can get this by solving for slope intercept form and then taking the opposite and recipricol of the slope. Solve the rest using point-slope form.

5 0

3 years ago

Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En

ArbitrLikvidat [17]

Y₁ and Y₂ are independent, so their joint density is

$f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}$