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2 years ago

Please answer! I’ll give you Brainliest

Mathematics

1 answer:

Sergio [31]2 years ago

4 0

The four small squares are all 3x3, so the perimeter is 3+3+3+3 = 12.

12 * 4 = 48

We can find the perimeter of the large diamond shape using the Pythagorean theorem, a^2 + b^2 = c^2

a = 4

b = 4

4^2 + 4^2 = c^2

16 + 16 = c^2

32 = c^2

c = 5.65685425

So the perimeter of the large diamond shape is

5.65685425+5.65685425+5.65685425+5.65685425 = 22.627417

48+22.6 = 70.6

The areas of the small squares are 3x3 = 9

9 * 4 = 36

The area of the large diamond shape is

5.65685425 * 5.65685425 = 32

36 + 32 = 68

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Which of the following shows the answer to the problem 3/13 x 5/21 = _____ in its simplest form?

Nikolay [14]

$\frac{3}{13} * \frac{5}{21}$
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3 years ago

Yancy has a board 25 feet long. He wants to cut the board into 4 1/2 foot lengths. Into how many 4 1/2 foot lengths can he cut i

seropon [69]

So first we would convert 4 1/2 foot lengths into a decimal.

4.5 feet

Now we divide 25 by 4.5

25 ÷ 4.5 = 5.5

So Yancy can cut the board into five 4 1/2 foot lengths because the question says how many can he cut.

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3 years ago

A recent national survey found that high school students watched an average (mean) of 7.8 movies per month with a population sta

notka56 [123]

Answer:

Step-by-step explanation:

Given that :

Mean = 7.8

Standard deviation = 0.5

sample size = 30

Sample mean = 7.3 5.4772

The null and the alternative hypothesis is as follows;

$\mathbf{ H_o: \mu \geq 7.8}$

$\mathbf{ H_1: \mu < 7.8}$

The test statistics can be computed as :

$z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z = \dfrac{7.3- 7.8}{\dfrac{0.5}{\sqrt{30}}}$

$z = \dfrac{-0.5}{\dfrac{0.5}{5.4772}}$

$z = - 5.4772$

The p-value at 0.05 significance level is:

p-value = 1- P( Z < -5.4772)

p value = 0.00001

Decision Rule:

The decision rule is to reject the null hypothesis if p value is less than 0.05

Conclusion:

At the 0.05 significance level, there is sufficient information to reject the null hypothesis. Therefore ,we conclude that college students watch fewer movies a month than high school students.

8 0

4 years ago

Hope u can see this I will give 10 points to start of sorry for pushing u guys to type

Kruka [31]

Right triangles. There is a 90 degree angle on both

6 0

4 years ago

Read 2 more answers

Two car washes charge a basic fee plus a fee based on the number of extras that

mezya [45]

Answer:

a. The first car wash charges more for the basic fee

b. 6 extras must be chosen for the total cost to be the same

Explanation:

The complete question is shown in the attached diagram

The general form of the linear equation is:

y = mx + c where m is the slope and c is the y-intercept

The y-intercept represents the initial value which, in the given problem, is the basic fee

For the first car wash, we are given the equation:

y₁ = x + 9

Now, we will start by getting the equation for the second car wash:

1- getting the slope:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{12-9}{4-2}=1.5$

The equation now is: y = 1.5x+c

2- getting the y-intercept:

To get the y-intercept, we will simply choose a point, substitute with it in the equation from 1 and solve for c. I will use point (2,9)

y = 1.5 x + c

9 = 1.5 (2) + c ............> c = 6

The equation for the second car wash is:

y₂ = 1.5x + 6

Part 1:

As mentioned above, the y-intercept represents the basic fee

For the first car wash:

y₁ = x + 9 ................> The basic fee is $9

For the second car wash:

y₂ = 1.5x + 6 ..............> The basic fee is $6

It is clear that the basic fee for the first car wash charges more for the basic fee

Part 2:

In both equations, x represents the number of extras. We want the total cost to be the same. So, we will simply equate y₁ and y₂ and solve for x

x + 9 = 1.5x + 6

1.5x - x = 9 - 6

0.5x = 3

x = 6

Let's verify:

For x = 6:

y₁ = x + 9 = 6 + 9 = $15

y₂ = 1.5x + 6 = 1.5(6) + 6 = $15

Total cost is the same. This means that 6 extras must be chosen for the total cost to be the same

Hope this helps :)

6 0

3 years ago