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3 years ago

Correct the mistake

Mathematics

1 answer:

ankoles [38]3 years ago

3 0

(12+8)\ 5 + 15=19... this solution works because 12+8=20.. 20/5=4 and then 4+15=19

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6. A Two-story building measures 21 root

KengaRu [80]

Answer:

1008 in

Step-by-step explanation:

84x12=1008

84 feet times 12 inches because 12 inches is a foot.

so 84 feet times 12 inches gets your answer. 84 times 12=1008

make sense?

7 0

3 years ago

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4. Yearly, students are invited to membership in the honor society for your major if their

sergeinik [125]

You will not invited

Mean is best measure of average.

Given:

Class rank is at least = 90th percentile

Rank 30 out of 128 student

Find:

1. Invited for membership or not

2. Best measure of average

Computation:

Rank in total = [(100 - Percentile)(Total students)]/100

30 = [(100 - Percentile)(128)]/100

3000 = [(100 - Percentile)(128)]

23.4375 = 100 - Percentile

Percentile = 100 - 23.4375

Percentile = 76.5625

Since Percentile smaller than 90, So, you will not invited to membership

Mean is best measure of average.

Learn more:

brainly.com/question/6033034?referrer=searchResults

3 0

3 years ago

Evaluate 14 - 2xy when x = -6 and y = 2.

valkas [14]

Answer: 38

Step-by-step explanation:

14 - 2xy

(substitute x and y)

14 - 2 (-6) (2)

14 + 24

8 0

3 years ago

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The math question is on the image<br>find the nth term of the sequence

Rus_ich [418]

Notice how Pattern 2 is Pattern 1 with 4 balls added in the bottom row.

Pattern 3 is Pattern 2 with 5 more balls.

Pattern 4 is Pattern 3 with 6 more balls.

Generalizing the trend, we expect Pattern $n$ to be identical to Pattern $n-1$ with $n+2$ more balls.

If $b_n$ is the number of balls in the $n$ -th pattern, then we have the recursive relation

$\begin{cases} b_1 = 6 \\ b_n = b_{n-1} + n + 2 & \text{for } n>1 \end{cases}$

We can solve this recurrence by substitution. Using the definition of $b_n$ , we have

$b_{n-1} = b_{n-2} + (n-1) + 2 \\\\ \implies b_n = (b_{n-2} + (n-1) + 2) + n + 2 \\\\ \implies b_n = b_{n-2} + 2\times2 + \bigg(n + (n-1)\bigg)$

$b_{n-2} = b_{n-3} + (n-2) + 2 \\\\ \implies b_n = (b_{n-3} + (n-2) + 2) + 2\times 2 + \bigg(n + (n-1)\bigg) \\\\ \implies b_n = b_{n-3} + 3\times2 + \bigg(n + (n-1) + (n-2)\bigg)$

and so on, down to

$b_n = b_1 + (n-1)\times2 + \bigg(n + (n-1) + (n-2) + \cdots + 2\bigg)$

Recall that

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then we find

$\displaystyle b_n = 6 + 2(n-1) + \sum_{i=2}^n i$

$\displaystyle b_n = 2n + 4 + \left(\sum_{i=1}^n i - 1\right)$

$\displaystyle b_n = 2n + 3 + \frac{n(n+1)}2$

$\displaystyle \boxed{b_n = \frac{n^2+5n+6}2}$

8 0

2 years ago

Which of the following choices is the variance of the sample shown here? 35 37 42 57 61 70

viktelen [127]

Answer:

205.5. should be it!

3 0

3 years ago

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