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3 years ago

Hiii please help! i’ll give brainliest if you give a correct answer tysm!

Mathematics

2 answers:

castortr0y [4]3 years ago

7 0

Answer:

12/25

Step-by-step explanation:

so what you would do is keep the first fraction (2/5) then change the symbol then flip the last fraction (5/6). so then you would multiply 2*6= 12 and 5*5=25. so the answer would be 12/25.

adelina 88 [10]3 years ago

5 0

Answer:

The answer is D

Step-by-step explanation:

You cross multiply 5 and 5 and 2 and 6 to create a new fraction

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If the circumference of a circle is x, then what is the area of the circle in terms of x?

grandymaker [24]

Answer:

x^2 / 4pi

Step-by-step explanation:

The area of a circle is (pi x r)^2

The circumference of a circle is 2pi x r

The find the radius, you need to divide x by 2pi, so x / 2pi

Then just plug it in as the radius:

(pi x (x / 2pi))^2 = x^2 / 4pi

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3 years ago

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3 years ago

Lucas wants to use the box with the greatest volume. Which shipping box should he use? Box_

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Answer:

He will use the D box because it has a volume of

$27 {ft}^{3}$

Step-by-step explanation:

$a)v = whl \\ = 1 \times 5 \times 3 \\ = 15 {ft}^{3}$

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hope this helps

brainliest appreciated

good luck! have a nice day!!

8 0

2 years ago

Read 2 more answers

Suppose that the ages of members in a large billiards league have a known standard deviation of σ = 12 σ=12sigma, equals, 12 yea

asambeis [7]

Answer:

$n\geq 23$

Step-by-step explanation:

-For a known standard deviation, the sample size for a desired margin of error is calculated using the formula:

$n\geq (\frac{z\sigma}{ME})^2$

Where:

$\sigma$ is the standard deviation
$ME$ is the desired margin of error.

We substitute our given values to calculate the sample size:

$n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23$

Hence, the smallest desired sample size is 23

3 0

2 years ago

for any positive integer n, the sum of the first n positive integers equals n(n 1)/2. what is the sum of all the even integers b

marusya05 [52]

The sum of all the even integers between 99 and 301 is 20200

To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.

In this case, the first even integer will be 100 and the last even integer will be 300.

nth term of the AP = first term + (n-1) x common difference

⇒ 300 = 100 + (n-1) x 2

Therefore, n = (200 + 2 )/2 = 101

That is, there are 101 even integers between 99 and 301.

Sum of the 'n' terms in an AP = n/2 ( first term + last term)

= 101/2 (300+100)

= 20200

Thus sum of all the even integers between 99 and 301 = 20200

Learn more about arithmetic progressions at brainly.com/question/24592110

#SPJ4

7 0

1 year ago