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3 years ago

Can someone answer this please!!

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

The perimeter of the square is option E. 16√5 mm.

The area of the square is option A. 80 mm².

Step-by-step explanation:

1. Let's calculate the perimeter of this square:

The perimeter of a square is the sum of its side 4 times, thus:

Perimeter = 4√5 + 4√5 + 4√5 + 4√5

Perimeter = 16 √5

Perimeter = 4 * 4√5 = 16√5

The correct answer is option E. 16√5 mm

2. Let's calculate the area of this square:

The area of a square is the length of its side squared, thus:

Area = (4√5)²

Area = 4√5 * 4√5

Area = (4 * 4 * √5 * √5)

Area = 16 * 5 = 80

The correct answer is A. 80 mm²

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Answer:

GCF = 5

Step-by-step explanation:

Given:

The numbers are given as:

10, 30 and 45

A factor of a number is a number by which the given number is evenly divisible.

Example: 2 is a factor of 4 as $4\div 2 =2$ . So, 4 is evenly divisible by 2 and hence 2 is a factor of 4.

Now, factors of 10 = 1, 2, 5, 10.

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30.

Factors of 45 = 1, 3, 5, 9, 15, 45.

So, the greatest common factor in all the three numbers is 5.

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3 years ago

the sum of three numbers is 55. the second number is twice the first and the third number is five less than the second. What are

ArbitrLikvidat [17]

Let the first number be = x

Then the second number = 2x

The third number = 2x - 5

Their sum = 55

This can be written in an equation as =

x + 2x + 2x - 5 = 55

= x + 2x + 2x = 55 + 5 ( transposing -5 from LHS to RHS changes -5 to +5 )

= x + 2x + 2x = 60

= 5x = 60

= x = 60 ÷ 5 ( transposing ×5 from LHS to RHS changes ×5 to ÷5 )

= x = 12

The first number = x = 12

The second number = 2x = 2 × 20 = 24

The third number = 2x - 5 = 24 - 5 = 19

Therefore , the three numbers are 12 , 24 and 19 .

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3 years ago

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Step-by-step explanation:

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2 years ago

In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala

Maslowich

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen A B d = a - b d²

1 13.76 13.74 0.02 0.004

2 12.47 12.45 0.02 0.004

3 10.09 10.08 0.01 0.001

4 8.91 8.92 -0.01 0.001

5 13.57 13.54 0.03 0.009

6 12.74 12.75 -0.01 0.001

∑ 0.06 0.0173

d`= ∑d/n= 0.006/6= 0.001

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sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

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3 years ago