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3 years ago

Write the equation of a line in slope intercept form that contains the points (-1,2) and (3,4)

Mathematics

1 answer:

Oksanka [162]3 years ago

3 0

Y=1/2x-4
I think thats the answer

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There are a total of 12 boys in a chess club. If the ratio of boys to girls in the chess club is 3 to 5, what is

n200080 [17]

<h3>Answer: 32</h3>

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Explanation:

b = number of boys = 12
g = number of girls = unknown

The ratio of boys to girls is 3 to 5, which means,

b/g = 3/5

12/g = 3/5 ... plug in b = 12

12*5 = g*3 ... cross multiply

60 = 3g

3g = 60

g = 60/3 .... divide both sides by 3

g = 20 is the number of girls

b+g = 12+20 = 32 total students in the chess club.

6 0

3 years ago

How would the average rate of change over years 1 to 5 and years 6 to 10 be affected if the population increased at a rate of 8%

Sever21 [200]

Answer:

(A): As year increases the number of pikas reduces.

(B): As year increases the number of pikas increases as opposed to when the rate reduces.

Step-by-step explanation:

<em>See comment for complete question</em>

<u>Given</u>

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

<u>(A) WHEN THE RATE DECREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1-8\%)^x$

Convert % to decimal

$f(x) = 144(1-0.08)^x$

$f(x) = 144(0.92)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(0.92)^x$

$f(1) = 144*0.92^1 =144*0.92=132.48$

$f(5) = 144*0.92^5 =144*0.66=95.04$

$Rate = \frac{95.04 - 132.48 }{4}$

$Rate = \frac{-37.44}{4}$

$Rate = -9.36$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(0.92)^x$

$f(6) = 144*0.92^6 =144*0.61=87.84$

$f(10) = 144*0.92^{10} =144*0.43=61.92$

$Rate = \frac{61.92-87.84}{4}$

$Rate = \frac{-25.92}{4}$

$Rate = -6.48$

So, we have:

$Rate = -9.36$ for year 1 to 5

This means that the number of pikas reduces by 9.36 yearly

$Rate = -6.48$ for year 6 to 10

This means that the number of pikas reduces by 6.48 yearly

So, we can say that, as year increases the number of pikas reduces.

<u>(B) WHEN THE RATE INCREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1+8\%)^x$

Convert % to decimal

$f(x) = 144(1+0.08)^x$

$f(x) = 144(1.08)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(1.08)^x$

$f(1) = 144(1.08)^1 = 144*1.08= 155.52$

$f(5) = 144(1.08)^5 = 144*1.47= 211.68$

$Rate = \frac{211.68 - 155.52}{4}$

$Rate = \frac{56.16}{4}$

$Rate = 14.04$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(1.08)^x$

$f(6) = 144(1.08)^6 = 228.52$

$f(10) = 144(1.08)^{10} = 310.89$

$Rate = \frac{310.89-228.52}{4}$

$Rate = \frac{82.37}{4}$

$Rate = 20.59$

So, we have:

$Rate = 14.04$ for year 1 to 5

This means that the number of pikas increases by 14.04 yearly

$Rate = 20.59$ for year 6 to 10

This means that the number of pikas increases by 20.59 yearly

So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.

5 0

3 years ago

You are rolling a pair of dice what's is the probability of getting different numbers on both dice

padilas [110]

Answer:

$Pr = 0.833$

Step-by-step explanation:

Given

A roll of a pair of dice

Required

The probability of getting different numbers

First, list out the sample space;

We have:

$S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6),(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6),$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6),(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}$

The total is:

$Total = 36$

The outcomes of the different numbers are:

$Different = \{(1, 2) (1, 3) (1, 4) (1, 5) (1, 6),(2, 1) (2, 3) (2, 4) (2, 5) (2, 6),$

$(3, 1) (3, 2)(3, 4) (3, 5) (3, 6),(4, 1) (4, 2) (4, 3) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) \}$

The total is:

$Different = 30$

So, the probability of having a different outcome is:

$Pr = \frac{Different}{Total}$

$Pr = \frac{30}{36}$

$Pr = 0.833$

3 0

3 years ago

Please help with khan!

frez [133]

Answer:

Step-by-step explanation:

n+12=18.3

-12 -12

n=6.3

He subtracted incorrectly and got 5.7 as an answer instead of 6.3

-hope it helps

4 0

3 years ago

Read 2 more answers

Mr.Chen left $45,000 to his aunt when he died. This was 5% of his total worth. a) how much was Mr.Chen worth when he died? Show

marysya [2.9K]

A) Mr Chen was worth:

45,000/5 = 9,000

9,000 x 100 = 900,000

The answer is 900,00

You can also write this as an equation.

3 0

3 years ago