All categories

2 years ago

Find the slope of the line that passes through (-34, 81) and (32, 12).

Mathematics

1 answer:

Sauron [17]2 years ago

7 0

Answer:-23/22 or -1.045

Step-by-step explanation:

slope = rise/run

Slope =y2−y1x/2−x1 (The 2's and 1's here should be subscripts sorry)

slope= 12-81 (the y's) divided by 32-(-34)( the x's)

12-81/32-(-34) (Parentehsis just to emphisis the minus a negative number the second coordinate is -32)

slope= -69/66

simplfy to -23/22 or to decimal

slope = -1.045

You might be interested in

75 + 13 + ?? = 140 i dont know what it is.

mixer [17]

Answer:

Step-by-step explanation:

140 - 13 = 127

127 - 75 = 52

Check your work:

75 + 13 + 52

= 140

8 0

3 years ago

Read 2 more answers

Which statement below is true about the

In-s [12.5K]

Answer:

the correct answer is d

3 0

3 years ago

Consider a single span of the spinner. Which events are mutually exclusive? Select two options. Landing on a shaded portion and

Sidana [21]

Answer:

Landing on a 3 and on a shaded area

Step-by-step explanation:

Mutually exclusive means there is now way the two events can happen at the same time. This is the only option where it can either be one or the other

5 0

3 years ago

Read 2 more answers

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago

Show all work below to answer the following for a certain city that had a population of 150,000 in the year 2010. In 2020, the p

Gelneren [198K]

The constant rate of continuous growth, k, for this population is equal to 2.11935%. And the population will reach 250,000 people in 24.36 years.

For solving this question, you should apply the Population Growth Equation.

<h3>Population Growth Equation</h3>

The formula for the Population Growth Equation is:

$P_f=P_o*(1+\frac{R}{100} )^t$

Pf= future population

Po=initial population

r=growth rate

t= time (years)

STEP 1 - Find the constant rate of continuous growth, k, for this population.

For this exercise, you have:

Pf= future population= 185,000 in 2020.

Po=initial population =150,000 in 2010.

r=growth rate= ?

t= time (years)=2020-2010=10

Then,

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 185000=150000\cdot \left(1+\frac{R}{100}\:\right)^{10}\\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{185000}{150000} \\ \\ \left(1+\frac{R}{100}\right)^{10}=\frac{37}{30}\\ \\ R=100\sqrt[10]{\frac{37}{30}}-100=2.11935\%$

STEP 2 - Find the <em>t</em> for population 250,000 people.

$P_f=P_o*(1+\frac{R}{100} )^t\\ \\ 250000=150000\cdot \left(1+\frac{2.11935}{100}\:\right)^{10}\\ \\ \left(1+\frac{2.11935}{100}\right)^{10}=\frac{250000}{150000} \\ \\ \left(1+\frac{2.11935}{100}\right)^t=\frac{5}{3}\\ \\ t\ln \left(1+\frac{2.11935}{100}\right)=\ln \left(\frac{5}{3}\right)\\ \\ t=\frac{\ln \left(\frac{5}{3}\right)}{\ln \left(\frac{102.11935}{100}\right)}\\ \\ t=24.36$

Read more about the population growth equation here:

brainly.com/question/25630111

7 0

2 years ago