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2 years ago

What is the best estimate of the sum of the fractions?

Mathematics

1 answer:

hoa [83]2 years ago

8 0

Answer:

14 1/2

Step-by-step explanation:

Estimations only help us do some rough calculations. They don't give us correct answer but, again, they aren't supposed to.

When estimating sum of the fractions, we should estimate each fraction separately and then just add those estimations up.

For example, 31/4 is somewhere just below 8, but we'll estimate it to 8. Similarly, 17/3, is a bit less then 6, but close enough to be rounded up.

Finally, 8/18 is somewhere near 1/2, so we round it up to that value.

Now, we simply add up all these estimations:

8+6+1/2 = 14 1/2

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What percent of discount should be given in a doll costing Rs 180

wolverine [178]

Answer:

11.11%

Step-by-step explanation:

So the question is:

$180 * (x)% = $y

Therefore $180 - $y = $160

Now:

180 - 160 = 20

Therefore; $20 is the discount

Now per cent of discount:

20/180 * 100 = 11.111111111111111111111%

therefore: per cent of discount is 11.11% (rounded to two decimal place)

3 0

3 years ago

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Question 4

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

Given

See attachment for chart

Required

Number of off days

To do this, we simply calculate the expected value of the chart.

This is calculated as:

$E(x) = \sum x * f$

Where

x = days

f = chances

So, we have:

$E(x) = 0* 0.3 + 1 *0.2 + 2 * 0.25 + 3 * 0.2 + 4 * 0.02 + 5 * 0.01 + 6 * 0.01 + 7 * 0.01$

$E(x) = 1.56$

$E(x) \approx 2$

3 0

2 years ago

What is the area of a garden that measures 4mx4mx6mx4mx10mx8m

topjm [15]

The area of the garden is 120 square metre

<h2>Explanation:</h2>

The picture of the garden is shown below. In this problem, we have:

One square
Two rectangles

Recall that the area of a square is given by:

$A_{s}=L^2 \\ \\ A_{s}:Area \ of \ a \ square \\ L:Side \ of \ the \ square$

On the other hand, the area of a rectangle is given by:

$A_{r}=L_{1}\times L_{2} \\ \\ \\ A_{r}:Area \ of \ a \ rectangle \\ \\ L_{1}:Side \ 1 \ of \ the \ rectangle \\ \\ L_{2}:Side \ 2 \ of \ the \ rectangle$

Therefore:

FOR THE SQUARE:

$L=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{s}=4^2 \\ \\ \boxed{A_{s}=16m^2}$

FOR THE RECTANGLE 1:

$L_{1}=6m \\ L_{2}=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{1}}=6(4) \\ \\ \boxed{A_{r_{1}}=24m^2}$

FOR THE RECTANGLE 2:

$L_{1}=10m \\ L_{2}=8m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{2}}=10(8) \\ \\ \boxed{A_{r_{2}}=80m^2}$

Finally, the area of a garden is the sum of the three areas:

$A_{total}=A_{s}+A_{r_{1}}+A_{r_{2}} \\ \\ A_{total}=16m^2+24m^2+80m^2 \\ \\ A_{total}=16+24+80 \\ \\ \boxed{A_{total}=120m^2}$

<h2>Learn more:</h2>

Area of a pizza: brainly.com/question/12878495

#LearnWithBrainly

4 0

2 years ago

Various studies indicate that approximately 11% of the world's population is left handed. You think this number is actually high

Irina-Kira [14]

Answer:

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

d. z= 1.3322

Step-by-step explanation:

We formulate our hypothesis as

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

According to the given conditions

p`= 31/225= 0.1378

np`= 225 > 5

n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5

p = 0.4 x= 31 and n 225

c. Using the test statistic

z= p`- p / √pq/n

d. Putting the values

z= 0.1378- 0.11/ √0.11*0.89/225

z= 0.1378- 0.11/ √0.0979/225

z= 0.1378- 0.11/ 0.02085

z= 1.3322

at 5% significance level the z- value is ± 1.645 for one tailed test

The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept Ha : p >0.11 and conclude that the data indicates that the 11% of the world's population is left-handed.

The rejection region is attached.

The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.

which appears to be 0.95 and subtracting from 1 gives 0.04998

8 0

3 years ago

A pentagon has 5 congruent sides and each angle measures 108°.

Dafna11 [192]

I'd say A because all side are equal.

6 0

3 years ago

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