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3 years ago

The top student in a distance throwing competition are shown in the table how many yard did the winner of the competition

Mathematics

1 answer:

V125BC [204]3 years ago

6 0

To correctly and efficiently, I need you to show me a table.

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Any one help me with this plsss and ni links plss!!

Bas_tet [7]

Answer:

6. D.

7. F.

8. A.

9. B.

10. C.

Step-by-step explanation:

6. 9 + (12 - 10)

12 - 10 = 2

9 + 2 = 11

7. (20 - 15) x 2

20 - 15 = 5

5 x 2 = 10

8. 10 ÷ 5 + 7

10 ÷ 5 = 2

2 + 7 = 9

9. 6 + 2 x 3

2 x 3 = 6

6 + 6 = 12

10. (2 x 4) + 8

2 x 4 = 8

8 + 8 = 16

Let me know if this helps!

4 0

3 years ago

What is the answer to this <br>u(2u+5)-4(U+3)

Yakvenalex [24]

Answer:

its 2u^2-4u+5u-12

Step-by-step explanation:

7 0

3 years ago

What is the percentage of cacao in hot chocolate which is made of 480 grams of milk and 20 grams of cacao?

galina1969 [7]

Given that the hot chocolate is made of 480 grams of milk and 20 grams of cacao.

We need to determine the percentage of cacao in hot chocolate.

<u>The percentage of cacao:</u>

The total grams of hot chocolate can be determined by adding the total grams of milk and the total grams of cacao.

Total grams of hot chocolate = Grams of Milk + Grams of cacao

Total grams of hot chocolate = 480 + 20 = 500 grams.

Thus, the total grams of hot chocolate is 500 grams.

Now, we shall find the percentage of cacao.

The percentage of cacao is given by

$\frac{100}{500}\times 20$

Simplifying, the values, we get,

$\frac{2000}{500}=4\%$

Thus, the percentage of cacao is 4%

6 0

3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago

The students in a statistics class record how many movies they have watched in the previous month. The data are listed below. 1,

ElenaW [278]

Answer:

A Histogram is a diagram consisting of rectangles whose area is proportional to the frequency of a variable and whose width is equal to the class interval "

Here we take a width of 2 units on the bar graph i.e. we consider the interval 0-2, 2-4, 4-6 and so on.

Now we arrange our table in these intervals with the help of the given quantities.

Step-by-step explanation:

3 0

3 years ago