What is the range of the function graphed below?

A12.5 ft ladder is leaning against the side of a house. If the ladder is resting 12 feet off the ground, how far is the ladder f

Elanso [62]

Answer:

3.5

Step-by-step explanation:

This can be calculatdd by applyinthe g Pythagorean theorem

a^2 + b^2= c^2

12^2 + x^2= 12.5^2

144+ x^2= 156.25

x^2= 156.25-144

x^2= 12.25

x= √12.25

x= 3.5

Hence the distance between the ladder and the house is 3.5 ft

6 0

2 years ago

Lui deposited $3,500 into a savings account. The simple interest is 4%. How much interest will the account earn in 2 years?

Varvara68 [4.7K]

Answer:

$280

Step-by-step explanation:

Principal = $3,500

Time = 2 Years

Interest Rate = 4%

Interest = Principal x Time x Interest Rate

Interest = $3,500 x 2 x 0.04

Interest = $280

6 0

3 years ago

Nancy spent half of her allowance going to the movies. She washed the family car and earned 7 dollars. What is her weekly allowa

Sauron [17]

(17-7)*2=weekly allowance
so her weekly allowance is 20

6 0

3 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

If you have to apply 30 n of force on a crowbar to lift an object that weighs 330 n, what is the mechanical advantage of the cro

shutvik [7]

<span>The answer is: 2. 11 Crowbar is a Class A lever Therefore the mechanical advantage of the crowbar can be given by either: MA = Output distance/Input distance OR MA = Output force/Input force Since, the question gives only the force, we can use the second formula. MA = Output force/ Input Force = 330 N / 30 N = 11</span>

5 0

3 years ago