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3 years ago

What is the length of the hypotenuse of the triangle?

Mathematics

2 answers:

Colt1911 [192]3 years ago

6 0

Answer:

8.1 ft

Step-by-step explanation:

hyp² = side² + side²

hyp² = 7² + 4²

hyp² = 49 + 16

hyp² = 65

hyp = sqrt of 65

hyp = 8.1 ft

PLEASE DO MARK ME AS BRAINLIEST UWU 

aksik [14]3 years ago

5 0

Answer:

$\large\boxed{\sqrt{65}\ ft}$

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4ft,\ leg=7ft,\ hypotenuse=AB$

Substitute:

$AB^2=4^2+7^2\\\\AB^2=16+49\\\\AB^2=65\to AB=\sqrt{65}$

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What number is between 27 and 45

Aloiza [94]

It could be 28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43, or 44.

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3 years ago

Read 2 more answers

The area of the rectangular floor in Tamara's room is 95 5/6 square

White raven [17]

Area of rectangle is width × length
which means area of rectangular floor in Tamara's room = width of room × length of room
-> 95 ⅚ sq feet = 8 ⅓ feet × length of
->(95 + 5/6) sq feet = (8 + 1/3) feet × length of room
->575/6 sq feet = 25/3 feet × length of room
length of room = (575/6)/(25/3)
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= 23/2
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2 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

Josefina pain rent for 5 months. She paid a total of $2045. how much did she pay per month?

djverab [1.8K]

Answer:

$409 per month

Step-by-step explanation:

If josefina paid $2045 over a time period of 5 months, then we have to divide how much she paid ($2045) by 5 to find out how much she paid for rent in 1 month.

We can use the equation: 2045 ÷ 5

When we solve that equation, we get 409 which means that josefina paid $409 per month for rent.

(sorry if this is confusing but I hope it helped)

8 0

2 years ago

Which correctly gives the location of the point (–14, 3)? A. Quadrant I B. Quadrant II C. Quadrant III D. Quadrant IV

dedylja [7]

The answer is B. Quadrant II

Just remember this concept instead.

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3 years ago