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3 years ago

9

Ronald can read at a constant rate of p pages per minute. Which equation represents the relationship between the total number of

pages, N, that Ronald can read in m minutes?

2 answers:

ANEK [815]3 years ago

5 0

N=pm The number of pages equals the pages read per minute

EleoNora [17]3 years ago

5 0

You have to do N=pm because the amount of pages= the pages that are read per minute

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Answer:

Step-by-step explanation:

Using the formula

p +/- z* √(p(1-p) / n)

Where p is sample proportion = 3226/3957 = 0.8153, n = 3957 and z* is 2.576

0.8153 + (2.576 √(0.8153(1-0.8153) / 3957))

0.8153 + (2.576 √ (0.8153(0.1847) /3957)

0.8513 + (2.576 √(0.1572/3957)

0.8513 + (2.576 √0.00003974)

0.8513 + 2.576(0.0063)

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0.8513 - 0.016

= 0.8353 ~ 84%

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3 years ago

Factorize the following:<br>5(3a + b)² +6 (3a +b) - 8

zvonat [6]

Answer:

Step-by-step explanation:

Here, following formula will be used

(a+b)²= a²+b²+2ab

question

5(3a + b)² +6 (3a +b) - 8

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here, we have expanded the formula

=> 15a²+15b²/9+15x2ab/3+18a+6b-8

=> 15a²+5b²/3+10ab+18a+6b-8

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3 years ago

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3 years ago

Uhm mainly 11 and 12 but if u solve all of them I’ll give u brainliest

Dvinal [7]

Question 11)

Answer:

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Step-by-step explanation:

Given the expression

$\log _2\left(63\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)$

$=\log _2\left(\frac{63}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7$

$=\log _2\left(7\right)$

$=2.80$

Therefore,

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Question 12)

Answer:

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

Step-by-step explanation:

Given the expression

$\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)$

$=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45$

$=\log _2\left(45\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)$

$=\log _2\left(\frac{45}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5$

$=\log _2\left(5\right)$

$=2.32$

Therefore,

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

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3 years ago